LIBRARY OF CONGRESS. 



©^■ 4t -.-5io*iijriB|t f a 

Shelf .OX?.£4- 

TJNITED STATES OF AMERICA. 



A TREATISE 



ON 



PLANE AND SPHERICAL 



TKIGONOMETBY. 



E. MILLER, A.M., 

Professor or Mathematics and Astronomy in the 
University of Kansas. 





LEACH, SHEWELL, AND SANBORN, 
BOSTON AND NEW YORK. 



[ X % *\^\ 



-O 

^ 



^ 



Copyright, 1891, by 
E. MILLER. 



^ V"~ 



Typography by J. S. Cushing & Co., Boston. 



Presswork by Berwick & Smith, Boston. 



PREFACE. 



This Treatise on Trigonometry has been written for use primarily 
in the classes of the University of Kansas. 

Throughout its entire preparation constant reference has been 
made to the works of Serret, Lonchampt, Young, Airy, Hind, 
Beasely, Todhunter, Newcornb, Chauvenet, Olney, " Oliver, Wait, 
and Jones," Wheeler, Peirce, Loomis, and others. The source of 
the material is to some extent to be found in those authors. The 
matter and the methods of presentation are designed to enable the 
student to become thoroughly acquainted with the principles and 
applications of Trigonometry; and care has been taken to render 
the demonstrations of the fundamental propositions as clear and 
as concise as possible, without in the least affecting their logical 
accuracy. 

In this volume the theory of the science is based upon the 
analytic method, and every practical formula is illustrated by 
examples of numerical computation. The Sets of Examples given 
are believed to be sufficient for all practical purposes, furnishing 
abundance, as well as variety, of work. Answers are not given, 
in order that the student may test his work by other methods 
and formula. 

Special acknowledgments are due Mr. H. B. Newson, Assistant 

in Mathematics, for a careful review of the manuscript, and for 

suggestions made. 

E. MILLER, 

University of Kansas. 
Decexeek, 1891. 



TABLE OF CONTENTS. 



CHAPTER I. 



Definitions 

Measurement of Angles 

Angular Notation , 

The Sexagesimal Method 

The Centesimal Method 

Complementary Angles " „ 3 

Supplementary Angles 3 

Examples. Set I 3 

CHAPTER II. 

The Trigonometric Functions 5 

Properties and Relations of Trigonometric Functions 5 

Functional Values of Angles of 30°, 45°, 60° 8 

Functions of the Form arc-sin 6, arc-cos 0, arc-tan 6, etc 12 

Examples. Set II 13 



CHAPTER III. 

Right Triangles 16 

Isosceles Triangles 18 

Regular Polygons 19 

Area of Right Triangles 19 

Examples. Set III 19 

CHAPTER IV. 

Formulae for the Sum or Difference of Two Angles 23 

(1) Sine and Cosine of the Sum of Two Angles 23 

(2) Sine and Cosine of the Difference of Two Angles 24 

(3) Tangent and Cotangent of the Sum of Two Angles 24 

(4) Tangent and Cotangent of the Difference of Two Angles 25 

Formulae for the Sum of Three Angles 25 

Functions of Double Angles 27 

Functions of Half Angles 27 

Examples. Set IV 29 



TABLE OF CONTENTS. 



CHAPTER V. 

PAGE 



Relations between the Angles and Sides of Oblique Triangles 31 

Area of Triangles 36 

Circumscribed, Inscribed, and Escribed Circles 38 

Examples. Set V 40 

CHAPTER VI. 

Solution of Oblique Triangles 41 

Examples. Set VI 45 

CHAPTER VII. 

Spherical Trigonometry 49 

Definitions 49 

Eundamental Eormulse 51 

Formulae adapted to Logarithmic Computation 55 

Formulae for the Eunctions of Half Angles 56 

Formulae for the Functions of Half Sides 58 

Formulae of Delambre 60 

Napier's Analogies 60 



CHAPTER VIII. 

Solution of Spherical Right Triangles 61 

Napier's Circular Parts 61 

Napier's Rules for Circular Parts 62 

Isosceles and Quadrantal Triangles 65 

Examples. Set VII 65 



CHAPTER IX. 

Solution of Spherical Oblique Triangles 67 

Examples. Set VIII 73 



CHAPTER X. 

Area of Spherical Triangles 75 

Formula of l'Huillier 76 

Radii of Small Circles circumscribing a Spherical Triangle 78 

Radii of Small Circles inscribed within a Spherical Triangle 79 

Radii of Small Circles escribed upon the Sides of a Spherical Triangle 80 

Examples. Set IX , . 81 



VI TABLE OF CONTENTS. 



CHAPTER XL 

PAGE 

Applications of Spherical Trigonometry 82 

Definitions of Astronomical Terms 82 

Examples. Set X 88 



CHAPTER XII. 
Miscellaneous Examples. Set XI 



CHAPTER XIII. 

Trigonometric Tables 93 

Values of Sines and Tangents between 0° and J it 93 

The Construction of a Table of Sines and Cosines 96 

The Construction of a Table of Tangents and Cotangents 99 

Tables of Logarithms of Trigonometric Functions 99 

Definitions of Terms used 100 

Properties of Logarithms 100 



Appendix. Formulae 103 



TRIGONOMETRY. 

CHAPTER I. 
INTRODUCTION. 

1. Trigonometry is a branch, of Mathematics, and comprises all 
investigations relating to the numerical computation of angles and 
triangles. 

2. Plane Trigonometry treats of the solution of plane triangles. 
It also includes the investigation of all the relations of angles, 
constituting the Angular Analysis, or Analytical Trigonometry. 

3. Measurement of Angles. The unit of angular measurement is 
an angle of one degree. 

A degree is one-ninetieth of a right angle, or of a quadrant of a 
circle. 

Fractional parts of a degree are represented by minutes and 
seconds; thus, 

one minute = g 1 ^ of a degree ; 

one second = -^ of a minute. 

4. Symbols are used to designate degrees, minutes, and seconds. 
An angle or arc of 30 degrees, 35 minutes, 50 seconds, is written 
30° 35' 50". To compute angles or arcs by degrees, minutes, and 
seconds, is called the sexagesimal method, and is in common use. 

5. The centesimal method consists in dividing a right angle into 
100 equal parts, called grades; a grade into 100 equal parts, called 
minutes; and a minute into 100 equal parts, called seconds. 

6. The ordinary conception of an angle, that it must be less than 
two right angles, is sufficient for geometric purposes and the solu- 
tion of plane triangles and other rectilinear figures ; but Trigonom- 

1 



PLANE TRIGONOMETRY. 



etry shows that angular magnitudes admit of indefinite increase 
or diminution. That is to say, an angle or arc may have a value 
anywhere between and + oo, or and — co, in the same manner as 
linear, or any other kind of extension. 

7. In Art. 3, it is stated that the unit of angular measurement 
is an angle of one degree. For the solution of triangles, this 
method is to be preferred. But for certain theoretical purposes, 
another standard unit may be taken; namely, the angle at the centre 
of a circle whose measuring arc is equal in length to the radius. 

If m represents the circumference of a circle, and r the radius, 
then, by Geometry, 

m = 2 7rr ; from which, if r = 1, 

m = 27r, 
\ m — ir. 

Now, let ABCEG (Fig. 1) be a circle whose centre is at 0. 

Then, the circumference may be represented by 2tt\ the semicir- 

cumference, by rr; and the quad- 
rant ABC, by J-tt. Let A be a 
fixed point upon the circum- 
ference ; P, a point starting at 
A and moving in the direction 
of the upper arrow. The arcs 
AB, ABC, ABCD, etc., are, by 
common consent, positive arcs. 
The arc AB is the measure of 
the angle A OB; the arc ABC 
is the measure of the angle 
AOC; the arc ABCD is the 
measure of the angle AOD, etc. 
If the point P, however, should 

move in the direction of the lower arrow, then the arcs All, AUG, 

AHGF, etc., are negative arcs. 

8. The diameter AE (Fig. 1) is called the initial diameter, and 
the point A, the origin of arcs. 

The diameter CG, perpendicular to AE, is the secondary diameter, 
and the point C, the secondary origin. 

9. The arc AP has no value at the origin; at C, it is \tv\ at 
E, tt; at G, |- tt; and at A, or once round, it is 2tt. For n circum- 




INTRODUCTION. 6 

ferences, the arc has a value of 2mr. We may imagine the move- 
ment of the point P to continue indefinitely, so that it shall describe 
an arc composed of many circumferences. 

10. The assumption of Art. 7 furnishes an 
invariable unit of angular measurement. 

Let AB be the arc of a circle equal in 
length to the radius CA, the centre being 
at a 

Since angles at the centre of a circle are 
proportional to their subtending arcs, we have 

anode ACB • arc AB radius 1 




4 right angles circumference 2-n- x radius 2?r 
.-. anode ACB = 4 ri £ nt an g les = 2 right angles 

2iTT 7T 

= ™L= 180 ° = 57° 17' 44.8" 
7T 3.14159 

= the unit of angular measurement, whatever 
may be the length of the radius of the 
circle. 

11. Complementary angles or arcs. Two angles or arcs, both 
positive, or one positive and the other negative, are complementary 
when their sum equals a quadrant or \-k. The complementary- 
angle of 66° is 24°; of 125° is -35°; of -150° is 240°. 

12. Supplementary angles or arcs. Two angles or arcs are sup- 
plementary when their sum equals a semicircumference or w. The 
supplementary angle of 40° is 140°; of 150° is 30°; of 270° is -90°;, 
and of -190° is 370°. 

13. Examples. 

SET I. 

1. Construct the angles, 75°; 100°; 120°; 200°; - 75 a ; -100°; 
- 300°; 720°; 1080°; - 500°; - 630°; 1900°; and name the quadrant 
to which each belongs. 

2. How many degrees in ivr ? |tt? 3tt? 3|-tt? -5tt? -|tt? 

3. Express in terms of tt, the arcs 15°; 18°; 36°; 45°; 90°; 120°; 
150°; 75° 15' 15"; 270°; 360°; 1800°; - 1440°; - 85° 25' 19". 



4 PLANE TRIGONOMETRY. 

4. How many times is the unit arc contained in 90°? 360°? 
300°? fx? 5tt? 4ti7t? 

5. What is the complement of 48° 12'? 125° 15' 16"? -80°? 

-120°? 7T? -7T? |7T? 77 + 30°? 7T - 30° ? 

6. What is the supplement of 275° 18' ? 48° 12' ? - 50° ? - 180° ? 
tt? -n? |tt? 24tt? ttH-50 ? 1^-50°? 2 U7T ? 

7. The radius of a circle being 10 inches, what is the length 
of an arc of 5° 30'? 45°? 360°? tt? |tt? vr + the unit arc? 10 
quadrants ? 

8. The radii of two concentric circles are 5 feet and 10 feet 
respectively ; what is the difference in feet between an arc of 60° 
on the one and of 60° on the other? 

9. The radius of a circle being 10 inches, how many degrees in 
an arc of 6 inches ? Of 29 inches ? Of 31.416 inches ? 

10. If radius equals one foot, how many degrees, minutes, and 
seconds in an arc of .56 of a foot ? Of .275 of a foot ? Of .9 of 
a foot ? 

11. The radius of a circle being 5 feet, what is the difference in 
degrees, minutes, and seconds between two arcs, one of which is 
15 feet long and the other 12 feet ? 

12. How many degrees, minutes, and seconds in an arc of 57r? 
Of 6tt? Of 8iTT? Of 9tt? 



THE TRIGONOMETRIC FUNCTIONS. 



CHAPTER II. 

THE TRIGONOMETRIC FUNCTIONS. 

14. The terms used to designate the Trigonometric Functions or 
Ratios are the words sine, cosine, tangent, cotangent, secant, cosecant, 
versed-sine, and coversed-sine, which are written sin, cos, tan, cot, sec, 
cosec, vers, and covers. 

The investigation .of the properties and the relations of these 
functions constitutes the chief part of Trigonometry. 

The magnitude of an angle is independent of the lengths of 
the lines by which it is formed ; and accordingly, since the ratios 
of the sides of a triangle remain unaltered, the magnitude of an 
angle may be determined by means of the Trigonometric Ratios, 
of which there are six. 

A right triangle being used for this purpose, the definitions of 
the functions are 

1. The sine of an angle is the ratio of the opposite side to the 
hypothenuse. 

2. The cosine of an angle is the ratio of the adjacent side to 
the hypothenuse. 

3. The tangent of an angle is the ratio of the opposite side to the 
adjacent side. 

4. The cotangent of an angle is the ratio of the adjacent side 
to the opposite side. 

5. The secant of an angle is the ratio of the hypothenuse to the 
adjacent side. 

6. The cosecant of an angle is the ratio of the hypothenuse to 
the opposite side. 

The definitions of versed-sine and coversed-sine are 

7. The versed-sine equals the difference between unity and the 
cosine. 

8. The coversed-sine equals the difference between unity and 
the sine. 



PLANE TRIGONOMETRY. 



15. Let ABC (Fig. 2) be any right triangle, right-angled at C, 
whose sides are a, b, and c, respectively. 

Then 




Fig. 2. 



(1) 


sin^l 


a 


(2) 


cos .4 


_b 

~ c 


(3) 


tan A 


a 


(4) 


cot A 


_b 

~ a 


(5) 


sec J. 


c 



(6) cosec A =-, 

(7) vers A* = 1 — cos A, 

(8) covers A* = 1 — sin A. 



[i] 



16. Certain other relations are obtained from the foregoing for- 
mulae, as follows : 

(1) tan^lxcotJ. =-x- = l; .'. tscnA = , 

b a cot A 

(2) sin A X cosec A = - x - = 1 ; . *. sin A = , 

c a cosec A 

bo 1 

(3) cos A x sec A = -x- = l; .*. cos A — — — , I m 

c b sec A 



(4) tan4 = S = « + 2 

V b c c 



sin^L 



(5) cot J. 



cos A 

b_b_^_a _ cos A 

ace sin A 




B In Fig. b\ we have 



sin^l = -, sin£ = ^, tan^ = ^, 

c c b 

a b -r, a , a b 

cos A = -, cos B = -, cot A = -, 



tanJ5 = -, 



cot 5 = -, 

6 



sec A = -, sec .B = -, cosec ^4 = -, cosec B = -> 
b a a b 



* The vers and covers being easily found from the cos and sin, we shall not 
use them hereafter. 



THE TRIGONOMETRIC FUNCTIONS. 

Therefore, since A + B = 90°, we find 



(1) sin ^L = - = cos B, 

c 

(2) cos A = - = sin B, 

c 

(3) tanu4 = - = cotJ3, 

(4) cot^ = - = tanJB, 

a 

(5) sec ^4 = - = cosec B, 

b 



(6) cosec A = - = sec B. 



[3] 



18. Since in any right triangle, as ABC (Fig. b') 1 

bc 2 +ac 2 = ab 2 , 



then 



BC 2 AC 2 = 1 
AB 2 AB 2 



BC AC 

Since = sin A, and —— = cos^l, by [1], 

AB AB 



.-. sin 2 A + cos 2 A = 1. 



19. Using the equation BC 2 -f AG 7 = AB', derived from Fig. b', 



we obtain 



AB 



= 1 



BC' 



,2' 



AC AC 

sec 2 J.= 1 + tan 2 A 



20. Again, from the equation BC 2 + AC 2 = AB 2 , we obtain 



^.-1 + 



ACT 



BC BC- 

cosec 2 J. = 1 + cot 2 A 



8 



PLANE TRIGONOMETRY. 



21. Collecting the results of Arts. 18, 19, 20, and others that 
follow easily from the same, we shall obtain 



(1) sinM + cosM = 1, 



(2) 
(3) 
(4) 
(5) 

(6) 

(7) 

(8) 



1 -j-tan 2 J.= sec 2 A, 
1 -f cot 2 A = cosec 2 A, 



cos A = 
tan^4 = 


Vl — sin 2 ^., 
sin A 






Vl — sin 2 .4 


cot A = 
sec A — 


Vl — sin 2 A 


sin A 
1 


cosec A = 


Vl — sin 2 -4 
1 



sin^. 



w 



22. The relations already established for angles not exceeding 
a right angle, hold universally, whatever be the magnitude of an 
angle, and whether positive or negative. 

It must be observed from (1) of [4] that sin^L = ± Vl — cos 2 ^., 
or cosA=:±-\/l — sin 2 A from (4) of [4], has a double sign. In 
such cases the angle or arc is either positive or negative, and it 
is necessary to determine from given conditions in any particular 
case ivhich sign must be used. 



23. The functional values of the angles 30°, 45°, and 60°, are 
so often used, an application of the formulae already obtained will 
now be made to find those values. 

Let ABD (Fig. 3) be an equilateral triangle, divided into two 
B right triangles by BC, a perpendicular from 

the vertex B upon the base AD. 
In the triangle ABC, 






AC=\AD = \A 




= b =ic. 


The angle 


A = 60°, 


and the angle 


ABC =30°. 



THE TRIGONOMETRIC FUNCTIONS. 9 

By [1], C osA=4^=-=i- ••• h J [3],cos60°=i = sin30°. 
AB c 

By [4], sin^i=Vl-cos 2 ^=4V3. .-. by [3], sin60° = jV3= cos30°. 

By r21tanJ.= ^L4. ... by [3],tan60°= V3=cot30°. 

COS -4 

By r21,cot^=^^. .-. by [3], cot 60°= A: = tan30°. 

J L J sin A L V3 

By [21, secA= -X_. .-. by [3], sec 60°= 2 = cosec30°. 

J L J cos^ L J 

By r21,cosec-4 = — — .\ by [3],cosec60°=— = sec30°. 

J L J ' sin J. ' L J V3 



24. Let the right triangle ABC (Fig. 4) have the angles A 
and 5, each, equal to 45°. Then the sides AC and BC will be 
equal ; i.e. a = b. 

By Geometry, AC 2 + £C 2 = AB 2 , or 
& 2 +a 2 = c 2 , or 

2 a 2 = c 2 , or 
a 2 -, 
c 




« 



sin^l =-• .-. by [3], sin45°= iy2 = cos 45°. 

tan .4 = -• .-. by [3], tan 45°= 1 = cot 45°. 



sec^L=— L — .-. by T31. sec 45°= V2 = cosec45°. 
cos A 

25. We shall now investigate the Trigonometric Functions of 
all the quadrants of a circle. 

Draw through 0, the centre of the circle ABCD (Fig. 5), the 
diameters AC and BD, at right angles to each other. 

Let us take A as the origin of arcs, and those arcs that have 
the direction of AMBH • ••, etc., as positive arcs, and those whose 
direction is that of AM'DT •••, etc., as negative (Art. 7). 



10 



PLANE TRIGONOMETRY. 



"We shall further assume that all lines drawn in the direction 

of, or parallel to, OA and OB, are positive, and those that are drawn 

in the direction of, or parallel to, OG and OD, are negative. 

The arc A MB is the first quadrant ; ^£T(7, the second ; <7TX>, 

the third ; and DMA, the fourth. 

Designate by the positive or negative quantity which represents 

the variable arc whose extremity M may take upon the circum- 
ference all possible positions. Let 
fall the perpendiculars, MN on 
OA, and ML on OB. Prolong the 
radius OM until it meets AE in 
E. The line OL or its equal 
MN, with its proper sign, is the 
sine of the arc AM, or 0. The line 
AE is the tangent of the arc 0, 
and the line OE is the secant. 

MB, the complementary arc of 
AM, or 6, has for its sine the line 
OJV or its equal ML ; for its tan- 
gent BF, and its secant OF. We 
e j therefore name the sine, tangent, 
and secant, of the arc MB, the 
cosine, cotangent, and cosecant, of 

the arc J.1T, or 0. Now designating the arc AM, as before, by ; 

the arc ABHhj $'; the arc ^LBOT by 0"; and the arc ABCDM' by 

0'", we shall have the following : 






Arc 6. 


Arc 6'. 


Arc 8". 


Arc 6'". 


sin . . . 


+ MN 


+ HS 


-TR 


- M'P 


cos . . . 


+ ON 


-OS 


-OR 


+ OP 


tan . 


+ AE 


-AE' 


+ AE 


-AE' 


cot . . . 


+ BF 


-BG 


+ BF 


-BG 


sec . . . 


+ OE 


-OE' 


-OE 


+ OE' 


cosec . . . 


+ OF 


+ OG 


-OF 


-OG 



26. Therefore, the sine of an arc is a positive or negative quantity 
which measures the perpendicular let fall from the extremity of an arc 
upon the diameter which passes through the origin. 



THE TRIGONOMETRIC FUNCTIONS. 11 

The tangent of an arc is the positive or negative quantity which 
measures that portion of the tangent line drawn from the origin of 
the arc and terminated by the diameter which passes through the ex- 
tremity of the arc. 

The secant of an arc is the positive or negative quantity which 
measures that portion of the diameter prolonged that is comprised 
between the centre and the tangent of the arc. 

The cosine, cotangent, and cosecant may be described in the 
same way. 

27. We shall now examine in what manner the six trigonometric 
functions of an arc vary, when that arc varies from to +oo, and 
from to -co. If increases from to +-J-7T, the six functions 
remain positive. Sin increases from to +1, while passing through 
all intermediate values. Tan 6 increases from to + go, and sec# 
from 1 to -f-oo. 

The cosine, cotangent, and cosecant, on the contrary, decrease ; 
that is to say, the cos $ decreases from 1 to ; cot decreases from 
-foo to ; and cosec 6 decreases from x to + 1. 

If the arc increases from Jtt to it, then sin0 decreases from + 1 
to ; tan increases from -co to ; sec 9 increases from -co to — 1 ; 
cos decreases from to — 1 ; cot 6 decreases from to — co ; and 
cosec increases from -}- 1 to + co. 

If the arc increases from it to f ?r, then sin0 decreases from 
to — 1 ; tan 6 increases from to + co ; sec 6 decreases from 

— 1 to — x ; cos increases from — 1 to ; cot decreases from + oc 
to ; and cosec 6 increases from — co to — 1. 

If the arc increases from f tt to 2tt, then sin0 increases from* 

— 1 to ; tan 6 increases from -co to ; sec decreases from + co 
to -fl; cos increases from to + 1 ; cot decreases from to — x ; 
and cosec decreases from — 1 to — x. 

28. If the arc increases from 2 it to 4tt, or from 4?r to 677, or 
•••to 2mr, the six functions will have periodically the same values 
and in the same order. 

If to the arc any number of circumferences be added, we shall 
have, whatever be the value of 0, and denoting by n any entire 
positive or entire negative quantity, 



12 PLANE TEIGONOMETEY. 

sin (2 mr -f- 0) = sin 0, cos (2 wtt + 0) = cos 0, 

tan (2 7i7r + 0) = tan 0, cot (2 iitt + 0) = cot 0, 

sec (2 wtt -f 0) = sec 0, cosec (2 W7r + 0) = cosec 0. 

29. If increases from to — oo, then whatever be the value of 0, 
sin ( — 0) — — sin 0, cos ( — 0) = cos 0, 

tan(— 0) = — tan0, cot (— 0) = — cot0, 

sec (—(9)= sec0, cosec (— 0) = — cosec 0. 

30. If be any arc, then and tt + terminate at the extremities 
of the same diameter, and we shall have, 

sin (tt 4- 0) = — sin 0, cos (tt -f 0) = — cos 0, 

tan (tt + 0) — tan 0, cot (tt + 0) = cot 0, 

sec (tt + 0) = — sec 0, cosec (7r -f 0) — — cosec 0. 

31. If in the equations of Art. 30 we change to — 0, we shall 
have, 

sin (tt — 0) = sin 0, cos (?r — 0) = — cos 0, 

tan(7r — 0) = — tan0, cot (tt — 0) = — cot 0, 

sec (tt — 0) = — sec 0, cosec (tt — 0) = cosec 0. 

From which it follows that if tivo arcs are supplementary, their 
sines and their cosecants are equal and of the same sign, but their 
cosines, tangents, cotangents, and secants are equal and of contrary signs. 

From the two preceding groups of equations, denoting by n any 
entire quantity positive or negative, we shall have, 

sin[(2?i + l)7r±0]=Tsin0, cos[(2w 4-1)tt ± 0] = -cos0, 
tan(ft7r ± 0) = ±tan0, cot(?i7r ± 0) =±cot0, 

sec[(2?i + 1)tt ± 0] = — sec0, cosec[(2?i + l)7r±0] = qF cosec 0. 

32. It is very important to remark that each of the trigonometric 
functions of an arc takes all values of which it is susceptible 
in the indefinite variation of throughout an interval of two 
quadrants. 

33. To the functions x = sin 0, x = tan 0, x = cos 0, x = sec 0, 
x = cot 0, etc., corresponds another class of functions, which are 
usually written by the Germans and the French, = arc-sin x, 



THE TRIGONOMETRIC FUNCTIONS. 13 

6 = arc-tan x, = arc-cos a;, = arc-sec x, 6 = arc-cot x, etc., and by 
others = sin~ 1 a;, = tan -1 a;, 6=cos~ 1 x, 6=sec~ J x, 6 = cot~ 1 x, etc. 
The German and French method is to be preferred. 

It is easily seen that = arc-sin a;, = arc-tana;, etc., are not 
entirely determined, for they admit of an indefinite number of 
values for each value of x. The expressions arc-sin a;, arc-tan x, 
arc-cot x, and arc-coseccc, become completely determined if their 
values are constantly comprised between — \tt and -\-\tt\ and in 
like manner, arc-cos x and arc-sec x will be determined if their values 
are constantly comprised between and tt. With these restrictions, 
the expressions arc-sin x, arc-tan x, arc-cos x, arc-cot x, arc-sec x, and 
arc-coseca;, may be considered as functions of x. 

34. Examples. 

SET II. 

Note. — Throughout this book, unless otherwise stated, radius 
is taken equal to unity. 

I. Construct : 



sin (9 = |, 


sin 6 = 7r, 


6 = arc-sin \, 


cos = -f , 


cos 6 = it, 


6 = arc-tan f, 


tan = 1, 


sin 6 = — -J, 


6 = arc-cos ^ , 


sec = 5, 


tan0 = -A> 


6 = arc-sec 2, 


tan0 = oo, 


cot 6=6, 


6 = arc-cot 3. 



2. Determine the values of the Trigonometrical Eatios for an 
angle of 585°. Also for an angle of 690°. Also for an angle of 
930°. Also for an angle of 6420°. 

3. Find all the angles between and 900° which satisfy tan 6 = 1. 
Find all the angles between and 900° which satisfy cos 2 = %. 

4. Find the values of the other functions of 6, when 

sin = |, sin 135° =£V2, 

sec 6 = 1, cos 120° =~h 

cot 6 = -, tan 1440° =0, 

n 

tan0 = --J, cot 540° = oo, 

tan0 = - 3, sin(27r + 30°) = i, 

cos0 = -i cos(tt +90°) = 0, 

esc 6 = - 1, sec(27r -j- 90°) = oo. 



14 



PLANE TRIGONOMETRY. 



5. Given tan = sin ; find the angle 0. 

6. Given cos = tan ; find the angle 0. 

7. Given cosec 45°= sec (180° -0); find the angle 0. 

8. Given cot (m + n)0 = tan 0; find 0. 

9. Given sini0 = cos (90 - 5£0); find (9. 

10. Given 10 sin = 2 tan ; find cos 0, sin 0, and tan 0. 

11. Given 2 sin 2 = 3 cos ; find 0. 

12. Given sin 2 0- 2cos0 + i = 0; find (9. 

13. Giventan0-f cot0 = 2; find 0. 

14. Show that tan0 + cot = seo '* + cosec ' g . 

sec0 cosec0 

15. Find the valnes of that will satisfy sin cos = tan 0. 

16. Show that sin 4 + cos 4 = 1-2 sin 2 cos 2 0. 

17. When is sin + cos = 1 ? When equal to — 1 ? When is 
sin 2 = 1 ? cos 2 = 1 ? When is tan 2 + cot 2 = 2 ? 



18. Find the value of sin 



3tt 

12* 



Of sin(-47r). Of cos-. Of 



cos(— 6tt). 

19. Construct arc-tan 1 ; arc-cos ( — J) ; arc-tan ( — 5) ; arc- 
cosec(— 5). 

20. Trace the changes in the sign and value of cos — sin 6, as 
6 passes from to 2ir. Also of tan0 + cot0. 

21. Find all the angles between and — - which satisfy the 



relation cos 2 = ■£-. 




22. Find sin and cos 0, when 
a + b = j-c. In this example use Fig. 
6, in which the angle A is denoted 
by 6, and the angle C is a right 
angle. 



Fig. 6. 



23. Find the value of 0, when tan 60 = 1. 

24. Find sin 6, when sec -f- tan = 2. 



25. Show that 1 + sin = 



cos 2 
1 — sin i 



THE TRIGONOMETRIC FUNCTIONS. 15 

26. Letting n = 0, 1, 2, 3, 4, 5, 6, in succession, show that 

sin4n- = ; sin(4w + 2% = ; sin(4?i + 3)? = - 1 ; 

Z Z -i 

cos(4w + 2)- = -l; and tan(2?i + l)^ = oo. 

27. Show that sin (270°- 0) = - cos0. 

28. Show that cos (270°- 0) = - sin0. 

29. Show that cos (270°+ 0) = sin0. 

30. Show that sin (360° - 0) = - sin 0. 

31. Show that cos (.360° - 0) = cos0. 

32. Show that sec ° + cosec ° = 1 + cot 6 = tau * + 1 - 

sec0 — cosec0 1 — cot0 tan — 1 

33. From the table of natural functions, find the 

sin of 15°, cos of 25° 15' 18", 

sin of 18° 15', cos of 135° 25' 20", 

sin of 75° 10' 35", tan of 60° 55' 43". 

34. From the table of logarithmic functions, find the logarithmic 
sin, cos, tan, and cot of 

18° 18' 18", 100° 50', 

50° 0'20", 175° 14' 25", 

150° 15' 25", 75° 16' 40". 

35. From the table of natural functions, find the angle whose 
sin is .25256 ; cos, .78543 ; tan, 3.14156 ; and cot, .56789. 

36. From the table of logarithmic functions, find the angle whose 
logarithmic sin is 9.12345; cos, 9.34567; tan, 10.43216; and cot, 
10.23456. 



16 



PLANE TRIGONOMETRY. 



CHAPTER III. 



TRIANGLES AND POLYGONS. 



35. To solve a triangle is to calculate the numerical value of 
its unknown elements, when a sufficient number of parts are given. 

In every triangle there are three angles and three sides. 

In a right triangle, the right angle is always known ; and to 
solve such a triangle, either an oblique angle and a side or two sides 
must be given. 

When the given parts are the three angles of a triangle, the 
problem is indeterminate ; that is to say, an indefinite number of 
triangles may, each, satisfy the conditions. 

36. Case I. 

Given two sides of a right triangle ; to find the other parts. 

Let ABC (Fig. 7) be a triangle, right- 
angled at C. Let a and b be the given 
sides ; then, by Art. 15, we have 

tan BAC = % or, cot ABC = -. 
b b 

.-. * log tan BAC = log a -log b + 10, 
or log cot ABC = log a — log b + 10. 




By Art 15, sin BAC = --, 

c 



.*. log c = log a — log sin BAC + 10. 
BAG + ABC = 90°, and c 



BAC + ABC = 90°, and c = Va 2 + b\ 

Example. Given a = 15.5, and b = 18.2. Find the side c and 
the angles A and B. 

Solution by logarithms : 

(1) loga = 1.190332 
log& = 1.260071 
log tan A = 9.930261 = log cot B. 
.-. ^ = 40°25'9".6, and B = 49° 34' 50".4. 



* See Chapter XIII. 



TEIANGLES AND POLYGONS. 17 

(2) logo = 1.190332 

log sin ^1 = 9.811828 
logc= 1.378504. 
.-. c= 23.905. 
Checks. 40°25'9".6+49 o 34'50".4 = 90°; and 



23.905 = V15.5 2 + 18.2 2 . 

37. Case II. 

Given the hypothenuse and one side; to find the other parts. 

Let c and a be given. 

We have, to determine the unknown parts, 




smi= cos±> = -, 
c 

and b 2 = c 2 — a 2 = (c + o) (c — o) . 

The direct determination of the angle B is 
obtained by the formula for the cosine ; but if the 
hypothenuse differs little from the given side, as 
is sometimes the case, then the angle B is not exactly determined. 
To avoid the difficulty, we may first obtain the side o, and then, by 

Art. 15, (3), [1], calculate the angle B from tanJ3 = -. 

a 

By logarithms : 

(1) log b = i [log(c + a) + log(c - a)], 

(2) logtan.B = log& — logo + 10, 

(3) log tan ^L = logo — logo + 10. 



Checks. A + B = 90°, and c = Vo 2 + b 2 . 

Example. Given c = 5892.51, and a = 5439.24. Find b, A, 
and B. 

Solution: c + o =11331.75, c-o = 453.27. 

By (1), log(c + o) = 4.054297 By (2), logo = 3.355327 

log(c - a) = 2.656357 logo = 3.735538 

2)6.710654 log tan B = 9.619789 

log b = 3.355327. .-. B = 22° 37' 11".4. 
.-. 6 = 2266.35. 



18 PLANE TRIGONOMETRY. 

By (3), loga= 3.735538 

log 6= 3.355327 

log tan i = 10.380211. 

.-. ^L = 67°22'48".6. 
Checks. 67°22'48".6-}-22 o 37'll".4 = 90 o ; and 



5892.51 = V 5439.24 + 2266.35 . 

38. Case III. 

Given an acute angle and one side; to find the other parts. 

Let the angle A and the side a be given. Then the required 
parts may be determined by the formulae 

(1) ^+5 = 90°; (2) sm^ = -; (3) tan^t = -. 

c b 

(2) and (3) changed to the form of logarithms become 
log c = log a — log sin A -f 10, 
log b = log a — log tan A + 10. 

39. Case IV. 

Given an acute angle and the hypothenuse; to find the other j)arts. 

Let the angle A and the hypothenuse c be given. Then the 
required parts may be determined by the formulae 

(1) ^ + £ = 90°; (2) sin^L = -; (3) cos-4 = -. 

c c 

(2) changed to the logarithmic form becomes 

log a = log sin A -f- log c — 10 ; 

(3) changed to the logarithmic form becomes 

log b = log cos A + log c — 10. 

40. Whatever may be the given parts of a right triangle, the 
required parts can always be found by the use of the formulae 



(1) c=Va 2 + 6 2 ; (2) A + B = 90°; (3) [1] and [3]. 

41. Isosceles Triangles. 

A perpendicular let fall from the vertex of an isosceles triangle 
upon the base divides the triangle into two equal right triangles. 
The triangle is solved by solving the right triangles. 



TRIANGLES AND POLYGONS. 



19 



42. Oblique Triangles. 

A perpendicular let fall from the vertex of an oblique triangle 
upon the side opposite divides the triangle into two right triangles. 
The oblique triangle is solved by solving the two right triangles. 

43. Regular Polygons. 

Let ABCDEF. (Fig. 8) be a regular polygon. 

Lines drawn from the centre to 
the middle points of the sides, as at 
G, are radii of the inscribed circle ; 
and lines drawn from the centre to 
the vertices of the polygon are radii 
of the circumscribed circle. In this 
manner a regular polygon is divided 
into twice as many right triangles as 
the polygon has sides. Let n be the 
number of sides of a regular poly- 
gon ; then the angle AOG will equal 
360^ = 180^ and the an le 0AQ wiu e al 9Q o_ 
2 n n 

If one of the two radii, OA or OG, or the side AB, be given, 
the angles having been determined as above, the remaining parts of 
the polygon may be found by the formulas for right triangles. 

44. Area of Right Triangles. 

The base b and the perpendicular a of a right triangle being 
given, the area, by Geometry, is equal to %ab. 

45. Examples. 










SET III. 




1. 


Given A = 50°, 




c = 15; 


to find B, a, b. 


2. 


Given A = 80°, 




6 = 40; 


to find B, a, c. 


3. 


Given A = 70° 


15', 


a = 225 ; 


to find B, b, c. 



4. Given A = 35° 35' 35", B = 54° 24' 25"; to find a, b, c. 

5. Given A = 59° 37' 42", c = 1785.395 ; to find B, a, b. 

6. Given B = 40° 45' 43", a = 15.15; to find A, b, c. 

7. Given B = 50° 50' 50", b = 201.356 ; to find A, a, c. 



20 PLANE TRIGONOMETRY. 

8. Given B = 75° 30' 38", c = 400 ; to find A, a, b. 

9. Given a = 65, b = 72 ; to find A, B, c. 

10. Given a = 2269, c = 3269 ; to find A, B, b. 

11. Given a = .00075, A = 75°; to find B, b, c. 

12. Given b = 99.5, c = 100 ; to find J, J5, a. 

13. The hypothenuse of a right triangle is 18 feet long, and one 
of the acute angles is 4 times the other. Find the angles and sides. 

14. In a right triangle whose hypothenuse is 15, and the angle 
A = arc-tan 3 ; what are the other parts ? 

15. In a right triangle the side a = 515.5 ; and the angle B, 
50° 45'. Find the other parts. 

16. The uniform grade of a railroad track 500 feet long is 4°. 
What is the elevation at the end ? 

17. What are the values of the trigonometric functions of a 
right triangle whose sides are 30, 40, 50, respectively ? Are there 
any other right triangles that give the same values ? 

18. The hypothenuse of a right triangle is 6 times the base. 
Find the angles. 

19. A tower 300 feet high casts a shadow 150 feet long upon 
the horizontal plane upon which it stands. Find the altitude of the 
sun. 

Isosceles Triangles. 

20. One of the two equal sides of an isosceles triangle is 18 feet 
long ; one of the equal angles, 30°. Find the third angle, the base, 
and the perpendicular from the third angle upon the base. 

21. Given the third angle of an isosceles triangle, and the per- 
pendicular from that angle upon the base. Find the equal angles, 
the equal sides, and the base. 

22. A chord 30 feet long is drawn in a circle whose diameter is 
60 feet. Find the angle at the centre. 

23. A circle whose diameter is 100 feet has two chords, one of 
which subtends an angle at the centre of 30°, and the other an angle 
of 50°. Find the difference in length between the two chords. 



TRIANGLES AND POLYGONS. 21 

Regular Polygons. 

24. Find the side of an equilateral triangle inscribed in a circle 
whose radius is 10. 

25. • Find the side of a regular decagon inscribed in a circle whose 
radius is 10. 

26. Find the radius of a circle in which is inscribed a regular 
pentagon whose perimeter is 50 feet. 

27. Find the radius of a circle in which is inscribed a regular 
dodecagon whose perimeter is 120 feet. 

28. Find the side of a regular pentadecagon inscribed in a circle 
whose radius is 50. 

Areas of Right Triangles and Regular Polygons. 

29. Find the area of a right triangle, the angle A being 75°, 
and the side a equal to 50. 

30. Find the area of a right triangle, A being equal to 50° 50', 
and c equal to 100. 

31. The area of a right triangle is 500, and the hypothenuse 200. 
Find the angles and sides. 

32. Find the area of a regular pentagon inscribed in a circle 
whose radius is 10 feet. 

33. Find the area of the space between a regular decagon in- 
scribed in, and a regular decagon circumscribed about, a circle whose 
radius is 10 feet. 

34. How much space between the circumference of a circle of 
20 feet diameter, and the perimeter of a regular dodecagon inscribed 
in the circle ? 

Miscellaneous. 

35. Solve the right triangle, when A == 30°, b = 100, and a = 40. 

36. ^4 = 18°, and c = 4 + V80. Find a, b, and B. 

37. Show whether there can be a right triangle when 

log a + 10 = log b -f log sin A. 



22 PLANE TRIGONOMETRY. 

38. If c cos 5 = b cos C, show that the triangle is isosceles ; that 
is, that b — c. 

39. The elevation of a tower is 30° to a man 6 feet high at a 
distance of 140 feet from the foot of the tower. Find the height of 
the tower. 

40. A man's shadow is twice his height. What is the altitude 
of the sun ? 

41. The sides of a right triangle are 20 and 32. Show that 
A = arc-tan |, B = arc-tan |, and the hypothenuse = 4 V89. 

42. Show that if 2 sin = tan 0, = 0, or 60°. 

43. If 6cot 2 6>-4cos 2 = l, then = ± 60°. 

44. If sin + cosec = 2, then = 90°. 

45. Show that sin /" 4n + 3 180° + o\=- cos 0. 

46. The difference of the lengths of the shadows of a vertical 
stick is 10 feet, when the sun's altitude is 45° and 30° respectively. 
Find the length of the stick. 

47. The difference of two acute angles of a right-angled triangle 
is 0°. Find the angles. 

48. Show that sin45 °~ sin30 ° = (sec 45°- tan 45 ) 2 . 

sin 45°+ sin 30° V ; 

49. The length of a kite string is 300 yards, and the elevation 
of the kite is 32°. Find its height. 

50. If a rectangle, 3 feet long by 2 feet broad, be taken as the 
unit of surface, what quantity will represent two square feet ? 



FORMULAE FOR ANGLES OB ARCS. 



23 



CHAPTER IV. 

FORMULA FOR THE SUM OR DIFFERENCE OF TWO 
ANGLES OR ARCS. 




46. To determine the sine and the cosine of the sum of two angles 
or arcs, the sines and the cosines of the angles or arcs being knoivn. 

Let AB and BC be two positive 
arcs whose sum does not exceed \ir. 
The origin of arcs (Fig. 9) being 
at A, represent the arc AB by 6, 
and BC by 0'. The arc BC may be 
considered as having its origin at 
B and its extremity at C. 

Then + 0'= the arc AC. 

Draw OF perpendicular to the 
radius OB, and CG, FH, BK, per- 
pendicular to OA, and FE parallel 
to OA. 

Then sin = BK, 

cos 6 = OK, 

sin0'= OF, 

cos0' = OF. 

(1) sm($ + $')=CG=CE + FH, 

(2) 008(0 + 0')= 0G= OH-EF. 

Radius is, in all these demonstrations, taken equal to unity. 
The similar triangles OBK and OFH give 

FH = 0H = OF 
BK OK OB 

From the first and third ratios we obtain 

(3) FH= BK o ° F = sin cos 0'. 



24 PLANE TRIGONOMETRY. 

From the second and third ratios we obtain 

(4) OH= 0K ' 0F = cos 6 cos 0'. 

v ' OB 

From the similar triangles OBK and CEF we have 

OK BK ~ OB 

.-. (5) CE = 0K ' GF =: cos OsinO'. 
v ; OB 

(6) FE = BK ' CF = sin sin 0'. 
OB 

Now, substitute (3) and (5) in (1), and (4) and (6) in (2); then 

sin (0 + 0') = sin cos 0'+ cos sin 0', [5] 

cos ((9 + 0') = cos cos 0'- sin sin 0'. [6] 

47. To determine the sine and the cosine of the difference of two 
angles or arcs, the sines and cosines of the angles or arcs being known. 

If in [5] and [6], 0' be changed to — 0', then 

sin (0 - 0') = sin cos 0' - cos sin $', [7] 

cos (0 - 0') = cos cos 0' + sin sin 0'. [8] 

48. To jfmd the tangent and cotangent of the sum of two angles 
or arcs. 

By Art. 16, (4), [2], 

V T ; cos (0 + 0') 
By [5] and [6], 

sin (0 + f ) = sin cos 0' + cos sin f 
cos (0 + 0') cos cos 0'— sin sin 0' 

Dividing both terms of the right member by cos cos 0', then 

Un(0 + e')= * axi$+ ^ ne ' . [9] 

v ; l-tan0tan0' L J 



FORMULAE FOR ANGLES OR ARCS. 25 

Again, by Art. 16, (5), [2], 

v ; sin (0 + 0') 

By [6] and [5], 

cos (0 + 0') _ cos cos f — sin sin 0' 
sin (0 + 0') ~~ sin cos 0' + cos sin 0' 

Dividing both terms of the right member by sin0 sin0', then 

cot (0 + 0') = <x>*0 cot 0'-l a [10] 

v } cot 0'+ cot L J 

49. 7b ^/md ^7ie tangent and cotangent of the difference of tivo 
angles or arcs. 

If in [9] and [10]; 0' be changed to -0', then 

tan(0-0') = tan 0- tan 0' -. 

v ) 1 + tan tan 0' L J 

cot(0-0') = 1 + co ^ cot ^ [12] 

v } cot0'-cot0 L J 



50. Formulae [5], [7], [9], and [11] have been demonstrated 
on the supposition that + 0' is not greater than \ic. It can be 
easily shown that they are generally true ; that is to say, that the 
sines, cosines, tangents, and cotangents of the sum of two angles or 
arcs, greater than \tt, iz, |-7r, 2ir, or 2mr, n being any finite integer, 
are all represented by the four formulae named. 

Tor example, 

sin(f tt+0+0') = sin(f 7T + 0)cos 0'+ cos(f tt + 0)sin0'. 

51. It is easily shown that 

sin(0 + 0'+ 0") = sin cos 0' cos 0"+ sin 0' cos cos 0" 

+ sin 0" cos cos 0' - sin sin 0' sin 0". [13] 

cos(0 + 0'+ 0") = cos0 cos 0' cos 0"- cos sin 0' sin 0" 

- cos 0' sin sin 0" - cos 0" sin sin r . [14] 



26 



PLANE TRIGONOMETRY. 



Knowing thus the formulae for the sine and cosine of the sum 
of three angles or arcs, we may obtain those which will give the 
sine and the cosine of the sum of four angles or arcs, or of five, or 
six, or of n angles or arcs. 

52. Important formulas deduced from [5], [6], [7], and [8]. 

From [5] and [6], sin(0 + 6') + sin(0 - 0') = 2 sin cos 0', " 
sin(0 +■ 0') - sin((9 - 0') = 2cos0 sin (9'. 

From [7] and [8], cos(0 + $') + cos(0 - 0') = 2cos0cos0', 
cos (0 + 6') - cos (0 - $') = - 2 sin sin $'. 

Now, let + 0' = <f>, and — $'= (f>[', then we shall have 

= «* + *')> and 0'= £(<*>- <*>')> 

and the preceding formulas, [15], become 



[15] 



(1) sin<£ + sin <£'=2sin£(<£4.<£')cos£(<£ -<£'), " 

(2) sin <£ - sin <f>'= 2sin£(<£ - <£')cos£(<£ + <£'), 

(3) cos<£ + cos<£'=2cos£(<£ + <£')cos£(<£ — <£'), 

(4) cos cf>' — cos <f> =2sin|-(<£ + <£')sin|(<£ — <£') 



!!' I- ™ 



These last formulae are frequently employed ; they serve to 
express the sum or difference of two sines or of two cosines in 
terms of the product of sines, or cosines, or of a sine and a cosine. 



53. We may express by a product the sum or the difference 
of a sine and of a cosine. 
We have 

cos cfi ± sin 4>'= sin(|-7r — <£) ± sin <£' ; 
and using [16], 

cos<£ + sinc/>'= 2sin/i,r - ±=£S cos^-iii^, 1 



( ^ _ ±±H\ C os (iir - ^=^ 



[17] 



FORMULA FOR ANGLES OR ARCS. 



27 



54. It is easily seen that, by division, a new set of important 
formulae may be derived from [16]. 



(i) 

(2) 
(3) 

(5) 
(6) 



sin <fr + sin<fr 



sin <f> — sin cf> 
sin<£-}- sin<£ 



cos <£ + cos <£ 
sin<£-fsin<£ 



cos</> — cos<£ 
sin (f> — sin <f> 



cos <f> -j- cos <f> 
sin<£ — sin<£ 



cos<£'— cos</> 



= sin£(<ft + <ft f ) cos £(<£-<£') = tan j-(<ft + <ft') 
sin !(</,-</,') cosi(</>+^) tani(^-^)' 



cos£(*+*') 

cos£(<fr — <£') 
sin £ (</>—<£') 

sin£(£— tf) 

cos£(<£ — <£') 

cos|(<fr + <ft') 
sin !<<£+<£') 



= tan*(++*')> 
= coti(*-*'), 
= tanj-(<£-0'), 

= cot* (* + *'), 



COS<fr+COS<fr r _ COS|-(<fr + <ft') COS|(<fr — <£') 

cos<£'— cos</> sin -j- (</>+<£') sin^(<^> — </>') 



[18] 



55. Functions of double angles or arcs. 

If in formulse [5], [6], [9], and [10], & be made equal to 6, then, 



sin 2 6 — 2 sin cos 0, 
cos 2 = cos 2 — sin 2 0, 
2 tan (9 



tan20 = 
cot20 = 



1-tan^' 
cot 2 0-l 



2cot0 

56. Functions of half angles or arcs. 
By [19], 2 sin J <9 cos ^0 = sin 0. 

By Art. 21, (1), [4], cos 2 -|0 + sin 2 i0 = 1; 
from which, by addition and subtraction, we obtain 
(cos£0 + sin|-0) 2 = 1 + sin0, 
(cos|0 - sin|-0) 2 = 1 - sin0. 



[19] 
[20] 

[21] 
[22] 



.\ cos^0 + sin-|-0 =±Vl + sin0, 
cos-j-0 — sin-J-0 = ±Vl — sin0: 



28 PLANE TRIGONOMETRY. 



sin-L0 = ± i( VI -f- sin0 ^ Vl- sin 6). [23] 



cos£0=±£(Vl + sin0 ±Vl-sin0). [24] 

Eesuming formulae [20], and (1) of [4], Art. 21, 
cos 2 i — sin 2 ^ = cos 0, 
cos 2 £0-f sin 2 i0=l, 
we obtain by addition and subtraction, and extracting the square root, 



cos 



sin 



ie = ± ^k±S!»°, [25 ] 

P = ±A /^p. [26] 



It will be noticed that [23] and [24] give four values for sin-J0, 
and four values for cos-J-0, respectively; and it is remarkable that 
the values of cos^-0 are precisely the same as those of sin-j-0. 

By (4) of [2], Art. 16, and [25] and [26], 



2 cos h Ml 



COS0 



i$ Al+cos<9 
which becomes, when both terms of the radical are multiplied by 



V 1 + cos 0, or by Vl — cos 0, 

tanj0 = sin * = l-<™g [27 ] 

2 1 + cos sin0 L J 

In like manner, cot|0 = 1 + cos ^ = sin ^ . [28] 

sin 1 — cos 

57. If we wish to express sin 20 or cos 20 in terms of sin0 
or cos only, we must replace sin by its value ± Vl — cos 2 0, and 
cos by ±Vl-sin 2 <9. 

Formulae [19] will then become 

sin20 = ± 2sin0Vl - sin 2 = ± 2cos0Vl - cos 2 0, [29] 
and [20] will become 

cos2<9 = 1 - 2sin 2 = 2cos 2 - 1. [30] 



FORMULAE FOE ANGLES OH ARCS. 29 

These results are such, that if we know the values of sin0 and 
cos0, then cos 2 will be completely determined, but the sin 20 will 
have two equal values, one of which is positive and the other nega- 
tive. It is important to remember the effect of the double sign, in 
order to determine the quadrant to which the function belongs. 

58. Examples. 

SET IV. 

1. Show that sin(45°-f 0)= cos(45 - 0). 

2. Show that sin 0= [ 



l+tan 2 i 



3. Show that cos fl + sin = tan 2 + sec 2 0. 

cos — sin 

4. Show that CQS (^ + 45 °) = sec 2 - tan 2 0. 

cos (0-45°) 

5. Find the value of in 90° + arc-sin = arc-tan 0. 

6. Show that sm ^ 0±e " > = GotO'±cotO. 

sin sin 0' 

7. Show that cos ( 0±6 ') = cot cot 0' q: 1. 

smO smO' 

8. Show that sinCfl + ^ tanfl + tan^ 

sin(O-O') tan 0- tan 0' 

9. Show that sin (^') = tanfl±W . 

cos (0:p0') 1 ±tan0tan0' 

10. Show that tan(0 -f 0'+ 0") 

__ tan0 + tan0'-ftan0"-tan0tan0 f tan0" < 
1 — tan tan 0' — tan tan 0" — tan 0' tan 0"' 

11. Show that sin 3 $ = 3 sin — 4 sin 3 0. 

12. Show that cos30 = 4cos 3 - 3cos0. 

13. Show that sin 4 = 4 (sin — 2 sin 3 0) cos 0. 

14. Show that sin 5 = 5 sin — 20 sin 3 + 16 sin 5 0. 

15. Show that tan 5 = tan40 + tan0 = tan30 + tan20 > 

l-tan40tan0 l-tan30tan20 



30 PLANE TRIGONOMETRY. 

16. Find the value of in 3tan0 tan30 + l = 0, when 0<-. 
Also, when < it. 

17. Show that 2 tan 2 = tan (45° + 0) - tan (45° - 0) . 

18. Show that c - ^- cos3 ^ tan2fl. 

sin 30 — sm0 

19. Find the value of when sin 40 + sin0 = 0. 

20. Find the value of when tan + tan (£tt + 0) = 2. 

21. Find tan when tan + ab cot = a + b. 

22. If + r + 0"= 180°, show that 

tan0 + tan0'+tau0"=tan0tan0'tan0". 

23. Take (10), and show that tan30 = 3tan ^~ tan ^ . 

v ; l-3tan 2 

24. If + 0'4- 0"= 90°, show that 

1 = tan tan 0'+ tan0 tan0"+ tan0' tan0". 

25. If + 0'+ 0"= 180°, show that 

sin0 + sin0'+sin0"=4cosJ0cos£0'cosJ0". 

26. Show that cos + cos (120°- 0) + cos (120°+ 0) = 0. 

27. Show that 4 sin sin (60° - 0) sin (60° + 0) = sin 3 0. 

28. Show that tan<9±tan ^ = ± tan tan 0'. 

cot ± cot 0' 

29. Find the value of sin (45°+ 30°). 

30. The functions of 30° being given (Art. 23), find those of 15°. 

31. Given sin = ra sin 0', and tan = n tan 0'; find sin and cos 0'. 

32. If + 0'+ 0" = 180°, show that 

sin 2 + sin 2 0'+sin 2 0"-2cos0cos0'cos0"=2. 

33. Given sin210°=~i; find cos 105°. 

34. Given tan 20 = — - 2 / ; find sin0 and cos0. 

35. Find the value of when 

sin30 + sin20 + sin0 = O. 



ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 



31 



CHAPTER V. 



ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 

59. Relations between the angles and the sides of an oblique triangle. 

Theorem I. In every plane triangle, the sides are proportional to 
the sines of the angles opposite. 

Let ABC (Fig. 10) be a triangle whose angles A and C are 
acute ; from the vertices B and C 
draw the perpendiculars BD to the 
base AC, and CE to the side AB. 

The two right triangles ABD and 
DBC will give, by Art. 15, 

BD — csin^l, 

and BD = a sin C. 

(1) .-. csinA = asinC. 




Fig. 10. 



The two right triangles CEA and CEB will, in like manner, 
give 



(2) 
Hence 



CE = b sin A, and CE = a sin B. 

.-. b sin A = a sin B. 

a b c 

sin A sinB sinO 



[31] 



If an angle of the triangle ABC (Fig. 11), C, for example, be 
obtuse, the perpendicular BD will fall upon 
the base produced. 

Since two supplementary angles have the 
same sine, the triangles ABD and CBD give, 
as before, 

BD = c sin A = a sin C ; 



from which we conclude that the preceding A 
formula is general. 




Fig. 11. 



32 



PLANE TRIGONOMETRY. 



We have, therefore, the three following relations between the 
angles and the sides of a triangle : 

A + B + = 180°, 
a b 



sinA sin 2?' 



[32] 



sin A sinC 

It is readily seen that these three formulae may be written 
ISO - B-C, 
a sin B 



A 
b 



sin (B + C)' 

— _a_smO__ 
€ ~sin(B + C)' 

60. Theorem II. In any plane triangle the square of any side 
is equal to the sum of the squares of the other two sides, less twice the 
product of these two sides by the cosine of their included angle. 

Let ABC (Fig. 12) be a triangle 
whose angle C is acute. Draw from 
the vertex B, to the base AC, the 
perpendicular BD. By Geometry, 
we have 

c 2 = a 2 + b 2 -2bxCD; 

but the right triangle BCD gives, 
C Art 15, 

CD = a cos C. 

.*. c 2 = a 2 + b 2 -2abcosG "1 
In like manner, a 2 = b 2 + c 2 — 2 be cos A, > [33] 

b 2 = a 2 + c 2 -2accosB. J 



If the angle C (Fig. 13) be obtuse, we 

have 

c 2 = a 2 + b 2 + 2bxCD. 

The right triangle CBD gives 

CD = a cos BCD. 





ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 33 

But cos BCD = cos (180° - BCA) = - cos BCA. 

.-. c 2 = a 2 + b 2 - 2 ab cos BCA. 
.-. the formulae [33] are general. 

61. If each of the angles A, B, and C be made successively a 
right angle, what form will [31] assume? What form will [33] 
assume ? 

62. Theorem III. In any plane triangle, any side is equal to 
the sum of the other tivo, each multiplied by the cosine of the angle 
which that side makes with the first side. 

Let BD be a perpendicular from the vertex B upon the base AC 
of the triangle ABC, (Fig. 14). 





Fig. 14. 



Fig. 15. 



We have, if the angles A and C are acute, 

b = AD + DC; 
but if one of the angles, as C, is obtuse (Fig. 15), then 

b = AD-DC. 
In the first case, DC= a cos C. 
In the second case, DC = b cos (180° — C) = — b cos C. 
And in both cases, AD = c cos A. 

We shall have 

(1) a = b cos C + c cos B, 

(2) b = c cos A 4- a cos C, 

(3) c = acosjB -h 6 cos A 



[34] 



34 PLANE TRIGONOMETRY. 

63. Formulae [34] may be obtained from [33] by adding those 
of [33] two and two, and reducing the results. Reciprocally, the 
first of [33] may be obtained from [34] by adding the three equa- 
tions of [34], after having multiplied (1) by a, (2) by b, and (3) 
by -c. 

In like manner the others may be obtained. 

64. Theorem IV. In any triangle, the sum of two sides is to 
their difference as the tangent of half the sum of the opposite angles is 
to the tangent of half their difference. 

From the fundamental formulae [31], 

sin A _ sin B _ sin C 
a b c 

we obtain 

m a + ^ — s i n ^- + sini ?_ 2sin ^ -(^ 1 + B) cos^ (A — B) 
c sin G sin C 

r>v a — b__ sin A — sin B _ 2sin^ (^1 — B) cos %(A + B) . 
{ } ~c~~ sinC " sin O ' 



[35] 



whence, dividing (1) by (2), we have 

a + b _ tan j- (A -f- B) 
a — b tan %(A — B) 

65. Other formulae relating to oblique triangles. 
The relation a 2 = b~ -f- c 2 — 2 be cos A, 

when changed to the form 

52 _j_ 2 _ a 2 

CO sA = -^— , 

26c 

will give the angle A; and the other two formulae of [33] will, in 
like manner, give the angles B and C. 

We shall now proceed to establish formulae for the functions of 
iA, \B, and \C. 

From [25] and [26] we have 



/i\ „ Anl ,1 /l + cos A 
(1) cosi^ = A/-^- o , 



(2) sin^ = ^i 



— cos A 



ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 



35 



7)2 I c 2 _ a 2 

and, substituting for cos A its value — — in (1), we have 

25c 

(1) cos|J. 

= \ 2bc + b 2 + c r ^~d 1 = / (6 + c) 2 -a 2 = / (a + & + c)(- a + b + c ) . 
\ " 46c \ 46c \ 46c 

and in (2) also, 



(2) sin £.4 



= / 25c-6 2 -c 2 + ft 2 = / a 2 - (6 - c)* = / (a + 6 - c)(a - 6 + c) 
"^V 46^ \ 46c \ 46c 



46c 

Now divide (2) by (1) ; 

then (3) ^-iA = t™lA=J (a + 6 -c)(a - 6 + <T 

cos \A \ ( a + 5 4. C ) ( _ a + 6 + c) 

If we put a + 6 + c = 2s, 

then — a-f-6 +c = 2(s — a), 

a - 6 + c = 2 (s - 6), 

a-h6 — c = 2 (s — c). 

Now substitute these values in equations (2), (1), and (3), and 
developing similar equations for \B and -J-0, we shall have three 
systems of formulae : 



* 6c 



sin 



Us-c) 



* ac 

* a6 

J* ac 
\ a,6 



[36] 



[37] 



36 



PLANE TRIGONOMETRY. 



2 \ s(s-a) 



o) 



tan ^3 = ^'-°) ('-") , 
2 V «(«-&) ' 

tanjg = J£= a > ( '- 6 > . 

2 \ s (s-c) 



[38] 



In all of [36], [37], and [38], the radicals must be taken 
with the sign + 5 for the half-angles of any triangle are, each, 
less than 90°, and consequently their trigonometric functions are 
positive. 

66. Area of Triangles. 

Let. the perpendicular BD be drawn from the vertex B of the 
triangle ABC to the base AC or AG produced. 





Fig. 17. 



Designating the area of a triangle by K, we have 
K=ibxBD. 

But from the right triangle ABD we have BD = c sin A, in 
either figure. 

.-. K=\bc sin A. [39] 

Therefore, the area of a triangle equals one-half the product of 
two sides multiplied by the sine of the angle included between those 
sides. 

Show that the area of any quadrilateral equals one-half the product 
of its diagonals multiplied by the sine of the angle between those 
diagonals. 



ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 37 

67. To find the area of a triangle in terms of its three sides. 

If in formula [39] we replace c by its value taken from [31], 

c sin C sinC whence c= & sin 



b sinB sin (A+C) sin (A + C)' 

rom -n -u tt i fr 2 sin^4sinO 

[391 will become K=f— — -■ 

siia.(A + C) 

Finally, from equations, 

. , A l(s-c)(s-b) 

and cos^ = J s(s ~ a) , 

* be 

established in Art. 65, we obtain 



a o • i a i a Vs(s — a)(s — b)(s — c) 

sini = 2sm£ A cos^A = 2— — * ^ *-± } - ; 

and if we replace sin ^4, in formula [39], by this value, then 



K=Vs(s-a)(s- b) (s-c). [40] 

By means of [36], [37], [38], and [40], it may be shown that 



(1) sin \ A sin \ B sin \ C = 



i n- (s-o)(s-b)(s-c) _ K 



abc sabc 



/o\ \ a i n i n s Vs(s — a)(s — b)(s — c) Ks I p/i 1 n 

(2) cos -J J. cos -J .B cos| = — - — * ^ ^ - = — , M 41 J 

abc abc 



(3) tan -J ^1 tan £ Stance = ~ 

s~ 



Show that (1) ! = eos^cos^ 

c sin -J- 

• 2\ s — c _ sin-|-^4 sin-|-jB 
^ ) c ~ sin^C ' 



(3) 



s — b _ sin j-^.cos-^-1? 
c cos -J (7 



f 4 ^ g Z a — C0S bA sin j-j? 
^ ) ~c~~ cos^O 



38 



PLANE TRIGONOMETRY. 



68. Radii of circumscribed, inscribed, and escribed circles. 

(1) Having circumscribed the circle whose centre is (Fig. 18) 

about the triangle ABC, draw from 
the vertex B the diameter BD — 2 It, 
and draw CD. The angle A = the 
angle D ; the triangle BCD is a right 
triangle. 

.-. a = 2RsinD = 2RsmA. 




Fig. 18. 



2 sin^t 

Multiplying both terms of ^—- — -7 
by be, we have 

n_ abc 



2 be sin A' 
then, multiplying both [39] and [40] by 4, we have 



4 K =2 be sin A == 4 V»(« — a) (s — 5) (s — c) . 

.*. it = - » 

4 Vs (s — a) (s — b) (s — c) 



[42] 



(2) Let the circle (Fig. 19), whose centre is and radius = r, 
be inscribed in the triangle ABC. The lines OA, OB, and OC 

divide the triangle into three tri- 
angles, whose bases are a, b, and c, 
respectively, and whose common alti- 
tude is r. 

As before, let 

K= the area of ABC, 

and 

jj then 




and 



s = J (a + b -f c) ; 



?* = 



But by [40] K= Vs(s - a)(s - b)(s - c). 

/ (g-q)(g-6)(g-c) 
A s 



.-. r 



[43] 



ANGLES, OBLIQUE TRIANGLES, AND CIRCLES. 



39 



(3) If we designate by r\ r ft , and r'" the radii of the escribed 
circles, that is to say, the circles which touch respectively the sides 




Fig. 20. 



a, 6, and c, and the prolongations of the other two sides, then it can 
be easily shown (Fig. 20) that from the area of the triangle ABC, 
which equals 

[40], Art. 67. 



we can obtain 



^s/s(s — a)(s — 6)(s — c) = K, 

= (s - a)r'= (s - b)r'"= (s - c)r", 



K 


-J S(S 


-b)(s- 


-o) 


s — a 


- \ 


s — a 


J 


K 


_J'(- 


-a)(s- 


-6) 


s — c 


~ \ 


s — c 


J 


K 


-J' s ^" 


— a)(s - 


-c) 



s-b 



s-b 



The formulse just obtained may, by Art. 65, be written 

r' = st&n%A, 1 
r" = s tan \ C, 
r"'=st2LniB. 



[44] 



[45] 



40 PLANE TRIGONOMETRY. 

From the foregoing formulae many others may be derived, among 
which are the following : 



1 1 



[46] 



4J?* = r'-H r"+r'"-r. 

Exercises. 

SET V. 

1. If a', b', c' be the perpendiculars from the centre of the cir- 
cumscribed circle upon the sides a, b, c, respectively, of the triangle 
ABC; then 

a , b , c __ 1 abc 
a' b' c'~l ' dVc 1 ' 

2. Prove that the product of the perpendiculars of a triangle 
from the angles on the opposite sides equals 

[(a + b + c)rj 
abc 
r being the radius of the inscribed circle. 

3. If B and r be radii of circles circumscribed about and inscribed 
in a regular polygon whose side is 2 a ; then 

B 2 -r 2 = a 2 . 

4. If r, r f be radii of the circumscribed and inscribed circles of 
a regular polygon of n sides, show that 

r + r r =acot— , 
2n 

where 2 a is one of the sides. 



5. If the radius of a circle be r, show that the length of an arc 

TrrO 

180* 
* R = the radius of the circumscribed circle. 



which subtends an angle of 0° at the centre is 



SOLUTION OF OBLIQUE TEIANGLES. 



41 



CHAPTER VI. 

SOLUTION OF OBLIQUE TRIANGLES. 

69. Case I. To solve a triangle, when two angles and one side 
are given. 

The unknown angle is obtained by the formula 

A + B + (7=180°. 
If a be the given side, we obtain the sides b and c by the formulae 

a sin B 





sin^l ' 




c = 


a sin C 

sin^l 




70. Example. Given A = 


:81°47'12".5, 




B = 


: 38° 12' 47".5, 




a — 


7012.24 ; 


to find C, b, and c. 


(1) A+B+C= 


180°. 


.-. C=60°. 


(2) 5 = asin5 . 
v J sinA 




/ON „ a sin (7 
( 3 ) c = • A ' 

SID. A 


loga = 3.845856 




loga = 3.845856 


log sin 5 = 9.791402 


l0( 


5 sin C =9.937530 


cologsin^L = 0.004477 


COlO£ 


j sin J. = 0.004477 


log 6 = 3.641735 


locc = 3.787863 


.-. 6 = 4382.65. 




.-. c = 6135.71. 



71. Case II. To solve a triangle, when two sides and the angle 
opposite one of them are given. 

If a, b, and A are the given elements, we may determine the 
unknown parts by the formulae 

• -d b sin A 
sinjB = , 



42 



PLANE TRIGONOMETRY. 



A + B + 0=180°, 
a sin C 



c = 



sin^. 



When the angle B differs but little from 90°, it cannot be exactly 
determined by means of the sine. 

In the case of a right triangle there can be no uncertainty as 
to the angles, because one being a right angle, either of the other 
two angles can be exactly determined from the formula A + B = 90°. 

But when an angle of an oblique triangle is determined from 
its sine or cosecant, then uncertainty may exist, since there are two 
angles each less than 180° having a given sine or cosecant. 

There can be no doubt as to the cosine, tangent, cotangent, or 
secant. 

In this case, the solution may determine two triangles, one 
triangle, or none. 

C 




Fig. 22. 



Suppose A to be the given angle, and a and b the given sides. 
The angle B is found by the formula 

gin _ B = 6sin^. 
a 
(1) If ^HL4<1 (Fig. 21), then two angles, B and B', will 

(Xi 

be determined, one > 90°, and the other < 90°. 

Now, if a > b, then the angle A > the angle B ; therefore B must 
be an acute angle, and there will be but one triangle. 

But if a<b, then either B or B' will meet the conditions, and 
two triangles will be determined. 

The angle C will have two values, and the side c also will have 
two values. 

b sin A 



(2) If 

triangle. 



= 1, it is clear that the figure is then a right 



SOLUTION OF OBLIQUE TRIANGLES. 43 

(3) if bsmA > i no triangle exists. 

a 

(4) If A>90° (Fig. 22), and a>b, there will be but one tri- 
angle, as in (1). 

But if A > 90° and a < b, there will be no triangle. 

Example. Given A = 27° 47' 44".77, 

a = 2199.12, 
6 = 2513.28 ; to find B, C, and c. 

(1) sin£=^A 
a 

•log 5 = 3.400240 

log sfai^L = 9.668685 

cologa = 4.657751 



log sin 5 = 1.726676 

.-. 5 = 32°12'15".23 
and 147° 47' 44". 77. 

(2) C= 180°- (A + B). 
.-.when B= 32° 12' 15".23, O = 120° 0' 0"; 

and when B = 147° 47' 44". 77, (7= 4° 24' 30".66. 

(3 ) c = ^^. 

sin^l 

log a = 3.342248 log a = 3.342248 

log sin C = 9.937530 log sin C = 8.885735 

colog sin^L = 0.331314 colog sin A = 0.331314 

logc = 3.611092 logc = 2.559297 

.-. c = 4084.08. .-. c = 362.493. 

72. Case III. To solve a triangle, when two sides and their 
included angle are given. 

Let a, b, and C be the given parts ; then, to find the angles A 
and B, we may employ formula [35], 

a + b = t&nl(A + B) 
a — b tani(J. — B)' 

and the side c by c = a _|HL^. 

SID. A 



44 PLANE TRIGONOMETRY. 

Example. Given a = 153, 

b = 137, 

= 40° 33' 12". 
(1) a + 6 = 290, a- b = 16, and \{A + B) = 69° 43' 24". 
log(a-b) = 1.204120 
log tani(^l + £) = 10.432446 
colog(a + &) = 3.537602 
logtani(^-^)= 9.174168 
.-. i(A-B)= 8° 29' 37", 
i(^L + 5) + i(^-5)=^ = 78 13' 1", 
i(^. + £) - \{A - B) = 5 = 61° 13' 47". 

(2) c = ^sin(? 
sin .4 
loga = 2.184691 
log sin C = 9.813018 
colog sin .4 = 0.0092 49 
logc = 2.006958 
.-. c = 101.616. 
Problems under this case may be readily solved by the formula 



c = Va 2 +b 2 -2abcos C, [33], Art. 60, 

when the side c is required. 

The angles A and B may then be found by [31]. 

73. To solve a triangle, ivhen the three sides are given. 

The formulae of [33], which determine the angles A, B, and C, 
by the sides, are not adapted to logarithmic computation ; but those 
of Art. 65 may be employed; particularly the last set, [38], which 
determine the angles \A, \B, and 1(7, by their tangents. 

Example. Given a = 701.224, 

b = 438.265, 

c = 613.571 ; to find A, B, and C. 

(1) Let s = i(a + 6 + c) =876.530, 

<s-a =175.306, 

s-b = 438.265, 

s-c = 262.959. 



SOLUTION OF OBLIQUE TRIANGLES. 



45 



(2) tan^l = J( s 



s — c) 
s(s — a) 

log(s- &) = 2.641736 

log(s-c) =2.419888 

colog (s- a) =3.756203 

colog $ = 3.057233 

2)1.875061 

log tani A =1.937530 

.-. i^=40°53'36 ,f .22. 
^=81°47'12".44. 



(3) ta^B =J(!=^E2 

log(s- a) = 2.243796 

log(s-c) =2.419888 

colog (s- b) = 3.358263 

colog s = 3.057233 

2)1.079180 



logtani£ = 1.539590 



IB 

i£ = 19° 6'23".77. 
J5 = 3S°12'47".54. 



(4) tan^=J (g - a)(S 7 6) . 

log(s- a) = 2.243796 

log(s- 6) = 2.641736 

colog(s-c) = 3.580111 

colog s = 3.057233 

2)1.522876 

log tan-j-C =1.761438 
.-. 1(7 = 30°, 
G = 60°. 



74. Examples. 



SET VI. 



1. The sides of a triangle are in the ratio of 2 : V6 : 1 + V3. 
Determine the angles. 

2. The sides of a triangle are 32, 40, and 66. Find the greatest 
angle. 

3. Given b = 14, c = 11, and A = 60°; to find 5. 

4. Given a = 70, b = 35, and C=36° 52' 12". Solve the triangle. 

5. Given a = 18, c = 2, and 5 = 55°. Find the other angles. 

6. A = 45° 50', jB = 60° 15', and c = 25. Solve the triangle. 

7. B = 25° 18', C = 130° 15', and b = 480. Solve the triangle. 



46 



PLANE TRIGONOMETRY. 



8. Two sides of a triangle are 3 feet and 5 feet respectively, 
and the included angle is 120°. Solve the triangle. 

9. The sides of a triangle are 4, 5, and 6. Find the angles. 

10. In a triangle ABC, the side AB is 254.3, the side AG 396.8, 
and the angle B 94° 21'. Solve the triangle. 

11. Given a = 201, b = 140, and B = 36° 44' ; to solve the triangle. 
Are there two triangles ? 

12. The sides of a triangle are 18, 19, and 20. Find the angles. 

13. Given A = 32°, a = 40, and b = 50 ; to solve the triangle. 
Are- there two triangles ? 

14. Given .4=18° 52' 13", a=27.465, and 6=13.189. Solve the 
triangle. 

15. The distance AB = 600 yards; the angle BAC=57° 35', 
and the angle ABC = 64° 51'. Find AC and BC 

1C. Find BC, the height of 
a hill above a horizontal plane ; 
AD being equal to 1000 feet; 
the angle A, 15° 36'; and ABB, 

27° 29'. 

17. A tower 150 feet high 
throws a shadow 75 feet long 
upon the horizontal plane on 
which it stands. Find the alti- 
tude of the sun. 

18. A person standing at the edge of a river observes that the 
top of a tower on the opposite edge makes an angle of 55° with a 
horizontal line drawn from his eye ; walking 30 feet back from the 
edge, the angle then is 48°. Find the breadth of the river. 

19. Find the area of a plane triangle whose sides are 24 feet, 
30 feet, and 18 feet, respectively. 

20. The sides of a triangle are 3, 5, and 6. Compare the radii 
of the circumscribed and inscribed circles. 

21. The sides of a triangle are 75 and 85, and the difference of 
the angles opposite those sides is 60°. Find all the angles. 




SOLUTION OF OBLIQUE TRIANGLES. 47 

22. Find the area of a triangular field, one of whose sides is 
45 rods long, and the two adjacent angles, respectively, 70° and 
69° 40'. 

23. How far off may a hill 500 feet high be seen, if the radius 
of the earth be 3962 miles ? 

24. Two fixed objects, A and B, and a ship, were all observed 
to be in a line bearing N". 33° 15' E. The ship then sailed north- 
west 10 miles, when the bearing of A was found to be east, and that 
of B northeast. Find the distance from A to B. 

25. The perimeter of a triangle is 400 feet, and the angles are 
65° 15', 75° 30', and 39° '15',. respectively. Find the sides. 

26. Express the area of a triangle, in terms of the base b and 
the two adjacent angles, 6 and $'. 

27. Two sides of a parallelogram are 90 feet and 110 feet, and 
one of the diagonals 120 feet. Eind the other diagonal, and the 
angles of the parallelogram. _ 

28. Given AB = 100 rods, 
angle BAD = 32°, 

BAC= 98°, 
ABO = 37°, 
ABD =118°; 

to find the distance between 
C and D. 

29. Eind the perpendiculars let fall from the vertices of a 
triangle upon the opposite sides, when a = 25, b = 30, c = 35. 

30. What is the angle of depression from the top of a mountain 
3 miles high, the earth's radius being 3962 miles ? 

31. A point of land was observed by a ship at sea to bear 
S. 11° 15' E. ; and after sailing northeast 12 miles, it was found to 
bear S. 33° 45' E. How far was the point from the ship at the 
last observation ? 

32. A side of the base of a square pyramid is 200 feet, and each 
edge is 150 feet. Eind the slope of each face. 

33. Erom the top of a mountain, 3 miles high, the angle of 
depression of the remotest visible point of the earth's surface is 




48 PLANE TRIGONOMETRY. 

2° 13' 27". Find the radius of the earth and the utmost distance 
from which the mountain is visible. 

34. Two sides of a triangle are 30.8 and 54.12, and the angle 
opposite the latter is 36° 42' 11". How many triangles ? Solve. 

35. Two sides of a triangle are 600 and 250, and the angle 
opposite the latter is 42° 12'. Solve the triangle. 

36. If cosfl= cosfl'-e ghow that tan+0 = \i±ftani0'. 

1 — e COS * 1 — e 

37. The angles of a triangle are as 3, 4, and 5, and the least 
side is 10. Find the other sides. 

38. The radius of the earth being 3962 miles, what is the length 
of 1° of the meridian ? 

39. At three points in the same horizontal straight line the 
angles of elevation of an object were found to be 36° 50', 21° 24', 
and 14°, the middle station being 34 feet from each of the others. 
Required the height of the object. 

40. If r, r', r", r'", denote the radii of the inscribed and escribed 
circles of a triangle, show that 

, 2 A rr' 
tarn— = — — • 



2 

41. Three circles, whose radii are r, r', r", touch one another ; 
show that the radius of the circle passing through the three points 
of contact is 

\ 



r + r' + r" 

42. The sides of a triangle are in arithmetical progression, and 
their common difference is 2 inches. If the area is 3 Vl5 square 
inches, find the sides. 

43. The area of a triangle is 84 square inches, and two of its 
sides are 15 and 13 inches. Find the third side. 

44. Given the vertical angle, the base, and the difference between 
the two sides of the triangle. Find the other angles. 



SPHERICAL TRIANGLES. 



49 



CHAPTER VII. 



SPHERICAL TRIANGLES. 



75. The object of Spherical Trigonometry is the solution of 
spherical triangles. 

The sides as well as the angles of a spherical triangle are 
expressed in degrees, minutes, and seconds. 

Only those spherical triangles whose sides are, each, less than 
180° will here be considered. 

76. Let ABC (Fig. 23) be a spherical triangle traced upon 
the surface of a sphere, whose centre is 0, and radii be drawn from 
O to the three vertices. A 
tri-edral angle will be formed 
whose plane angles AOB, 
AOG, and BOC, are respec- 
tively equal to the sides c, 
b, and a, since the latter are 
the arcs that measure those 
angles. 

The di-edral angles AOB- 
AOC,AOB-BOC, and AOC- 
BOC are equal to the angles 
A, B, and C, respectively. 
The planes which form the 
tri-edral angle AOB-AOC- 
BOC always cut the surface 
of the sphere in arcs of great circles, thus tracing upon the sur- 
face of the sphere a spherical triangle, whose sides are arcs of great 
circles. 




77. Definitions. 

A spherical right triangle has one, two, or three, right angles. 
When it has two right angles, the triangle is called bi-rec- 
tangular. 

When it has three right angles, the triangle is tri-rectangular. 



50 



SPHERICAL TRIGONOMETRY. 



A spherical triangle having one side = 90° is called quadrantal. 
When it has two sides, each = 90°, it is bi-quadrantal ; and when 
three sides, each = 90°, it is tri-quadrantal or tri-rectangutar. 

78. Relations between the angles and the sides of a spherical 
triangle. 

Let ABC (Fig. 24) be a spherical triangle traced on the surface 
of a sphere whose centre is O, and with a radius taken equal to 
unity. Let the sides b and c, each, be less than 90°. 




Fig. 24. 

Draw the tangents AD and AE, meeting the radii OB and OC 
prolonged, in the points D and E, respectively. 

We have AD = tan c, OD = sec c, AE = tan b, OE = sec b. 
Also, the angle DAE = A, and the angle DOE = a. 
The plane triangles DAE and DOE give 



(1) DE = AD + ^ - 2.4Z> -AE cos Z>A£, 

(2) DE 2 ='OD 2 + OE 2 - 2 OD-OE cos DOE; 

from which we obtain 

(3) 
20D-OE cos DOE=(OD 2 -AD 2 ) + (OE 2 ~AE 2 )-{-2AD'AE cosDAE. 

Since AD and AEJ are tangents at the point A, the extremity 
of the radius OA, the angles DAO and .EL40 are right angles, and 
(3) becomes 

(4) 2 OD • O^ cos DOE=20A 2 +2 AD -AE cos DAE. 



SPHERICAL TRIANGLES. 



51 



Eeplacing the various quantities in (4) by their values, we' have 

(5) sec b sec c cos a = 1 -}- tan b tan c cos A, 
whence multiplying (5) by cos b cose, we obtain 

(6) cos a = cos b cose -f- sin b sine cos A. 

79. We have supposed the sides b and c, each, less than 90°, 
but we shall now show that whatever values b and c may have 
between and 180°, (6) holds good. 

Suppose c > 90°, and b < 90°, and prolong the arcs BA and BC 
(Fig. 25) until they meet at jB'. b[ 
Let AB'= c', and CB'= a'-, then the 
triangle AB'C will give 

cosct'= cos& cose' 

-f sin 6 sin c' cos B'AC, 

for the sides c f and 5 are each less 
than 90°. 

Eeplacing a', c', and B 'AC by their values 180°— a, 180°— c, 
and 180°— A, we obtain as before 

cos a = cos b cos c + sin b sin c cos A 

Again, suppose b > 90° and c > 90°. 

Prolong the sides AB and ^4(7 (Fig. 26) until 
they meet at A'. Let <4'C = b', and J.'£ 
the triangle A'BC will give 




Fig. 25. 



c'; then 



cos a = cos b' cos c' -+- sin b' sin e'eos ^4'. 

Eeplacing b', c', and J/ by their values 180°— b, 
180°— c, and A, we obtain as before 

cos a = cos b cos c -f- sin b sin c cos A. 

We conclude, therefore, that whatever values the 
sides may have between the limits of and 180°, 
the formula holds. 

Thus by a change of letters we have three 
equations, 

cos a = cos b cos c + sin b sin c cos A, 

cos &= cos a cose + sin a sine cos B, > [47] 

cos c = cos a cos b + sin a sin b cos O, 

which may be regarded as the fundamental formulae of Spherical 
Trigonometry. 




52 



SPHERICAL TRIGONOMETRY. 



80. Relations existing between a side and the three angles of a 
spherical triangle. 

Let ABC be a spherical triangle whose polar is the triangle 
A'B'C (Fig. 27). 




From the geometric principles of polar triangles, we have 
A'=lS0°-a, a'=lS0°-A, 

B'=180°-b, b'=!S0°-B, 



C'=lS0°-c, 



i8o°- a 



By [47], 



cosa'= cos b' cos c'+ sin b' sin c' cos A'. 

Replacing the values of A', B', C, a', b', and c', we have 
cos (180°- A) = cos (180°- B) cos (180°- C) 

+ sin(180°- B) sin (180°- C) cos(180°-a), 
or — cos A = ( — cos B) ( — cos C) + sin B sin C{ — cos a) . 

This equation, after changing the signs, and two others similar 
to it, constitute the following group, 



cos A = — cos B cos C + sin B sin C cos a, 
cos £ = — cos A cos C -f- sin ^4 sin C cos 6, 
cos C = — cos yl cos JB +■ sin A sin _B cos c. 



[48] 



SPHERICAL TRIANGLES. 53 

81. Relations existing between two sides and the angles opposite. 

To obtain one relation between the sides a and b, and the angles 
A and B, is accomplished in a very simple manner by introducing 
into formulae [47] the sines of the angles A, B, and C, in place of 
the cosines. 

The first of [47] gives 

, cos a — cos b cos c 

cos A = ; — — 5 

sin b sm c 

from which 

sin 2 A = 1 - cos 2 ^ = sin2 fr sin2g ~ ( CQS a ~ CQS ft CQS C Y 

siirfr sin 2 c 

_ (1 — cos 2 &) (1 — cos 2 c) — (cos a — cos b cos c) 2 
sin '6 sin 2 e 

_ 1 — cos 2 a — cos 2 6 — cos 2 c + 2 cos a cos b cos c , 
sin 2 6 sin 2 c 

and, dividing both members by shra, we have 

sin 2 A _ 1 — cos 2 q — cos 2 6 — cos 2 c + 2 cos a cos b cos c 
sin 2 cc sin 2 a sin 2 6 sin 2 c 

This value of does not change when we permute the 

sirra 

letters a, b, and c. It follows, then, that the other two formulae of 

[47] will give the same value for and - — — , precisely as for 

. 2 sin-6 sm 2 c 

. • Therefore, since the angles and the sides of a triangle are, 

sin-a 

each, less than 180°, their sines are positive, and we have 

sin A _ sin B _ sin C 
sin a sin b sine 



[49] 



Therefore, in a spherical triangle the sines of the angles are pro- 
portional to the sines of the opposite sides. 

82. Relations existing between the three sides and two of the angles 
of a spherical triangle. 

If the value of the cose given in the third equation of [47] be 
substituted in the first of those equations, we shall have 

cos a = cos a cos 2 6 -f- sin a sin b cos b cos C + sin b sin c cos A, 



54 



SPHERICAL TRIGONOMETRY. 



which, by transposition, putting for cos 2 6 its value 1 — sin 2 &, and 
dividing by sin b, becomes 

cos a sin 6 — sin a cos b cos C = sin c cos A. 

By permutation of the letters, five other similar formulae may 
be obtained, so that we have 

cos a sin b — sin a cos b cos C = sin c cos -4, 
cos b sin a — sin b cos a cos C — sin c cos 5, 
cos 6 sin c — sin & cos c cos ^4 = sin a cos 5, 
cos c sin 6 — sin c cos 6 cos ^4 = sin a cos (7, 
cos c sin a — sin c cos a cos B = sin 6 cos C, 
cos a sine —sin a cose cosjB = sin& cos A. 



[50] 



sin a, sin b, and sine, by [49], may be replaced by sin A, sin B, 
and sin (7, and thus we shall obtain 



cos a sin B — cos & cos C sin ^4 = cos A sin (7, 
cos b sin ^4 — cos a cos (7 sin 5 = cos B sin (7, 
cos b sin (7 — cos c cos ^4 sin B — cos 5 sin^4, 
cos c sin jB — cos b cos .4 sin C = cos (7 sin A, 
cos c sin ^4 — cos q, cos jB sin C = cos (7 sin B, 
cos a sin C — cos c cos B sin ^4 = cos A sin 5. 



[51] 



83. Relations existing between tivo sides, the angle included between 
those two sides, and the angle opposite one of them. 

If we divide the first equation of [50] by 

sine sin A 



sin a = 



sin (7 



[49], 



member by member, the first of the relations sought will be found. 
We may obtain the others by permutation of the letters. 



cot a sin b — cot A sin C = cos b cos C, 
cot b sin a — cot B sin C = cos a cos (7, 
cot & sin c — cot _B sin A = cos c cos A, 
cot c sin b — cot sin A = cos 5 cos A, 
cot c sin a — cot C sin 5 = cos a cos 5, 
cot a sin c — cot A sin 5 = cos c cos 5. 



[52] 



SPHERICAL TRIANGLES. 



55 



84. Formulae relating to spherical right triangles, when there is 
but one right angle. 

If, in the first equation of [47], in two of the equations of [49], 
in the first, third, sixth, and fourth of [p2~], and in all of [48], 
each of which contains the angle A, we suppose the angle A equal 
to 90°, then the following formulae, peculiar to spherical right 
triangles, will result. 

From [47], (1) cos a = cos b cose, 
[49], (2) sin b = sin a sin B, 
(3) sin c = sin a sin C, 
[52],. (4) tan b = tan a cos C, 

(5) tan b = sin c tan B, 

(6) tan c = tan a cos B, 

(7) tane = sin b tan C, 
[48], (8) cosa = cotB cot C, 

(9) cos B— cos b sin C, 
(10) cos C= cose sin B. 

85. General formulae adapted to logarithmic computation. 

In order to render [47] and [48] suitable for logarithmic com- 
putation, let us resume the fundamental formulae [47], 

cos a = cos b cos c + sin b sin c cos A. 
cos a — cos b cose 



[53] 



(1) cos A = 

(2) l-cos^l = l- 

(3) l + cos^ = l + 



sin b sine 
cos a — cos b cos c 

sin 6 sine 
cos a — cos b cos c 



sin 6 sine 

But by substituting the values of 1 — cos ^4, and l+cos^4, as 
found by [26] and [27], (PL Tr.), we shall find that (2) becomes 



(4) smiA 



(sin 


b sine -f- cos 6 cose - 


- cosa 




|cos(6 — 
\ 2 sir 


2 sin b sin c 

c) — cosa 
tb sine 






_J sin 


i(« 


— 5 + c)sin^(a 


+ 5- 


o) 



sin b sine 



(Art. 52.) 



(Art. 65, PI. Tr..) 



fsin(s — 6)sin(.s — c) 
\ sin 6 sin c 



56 SPHERICAL TRIGONOMETRY. 

Operating upon (3) ma similar manner, we find 

X A — l C0S a ~ C0S ^ C0S C ~^~ sm ^ S * n C — 
\ 2 sin & sine 

| sin^(q + b + c)sin j-(fr -+- c — a) 
\ sin 6 sine 



(5) cos 



cos a — cos(&-i-c) 



2sin& sine 



sins sin(s — c) 



* sin b sine 
Now, if (4) be divided by (5), the result will be 



(6) taaM=J sin ( s - 6 > sin ^- c > - 
\ sin s sin (s — a) 

By a simple change of letters, analogous expressions for the sine, 
cosine, and tangent, of the angles \B and \ C, may be found. 
The three systems are as follows : 



• , . /sin(s — &)sin(s 
\ sm o sin c 



— 6)sm(s — c) 



. , -o /sm(s — a.)sm 

sin J I? =\ * — . — ^__ 

\ sm a sm 



c 



sm 



i p — /sin(g — a)sin (s — b) 
X sin a sin 6 



[54] 



cos J 



I _ / sins sin(s — a) 
\ sin 6 sine 

cosJ£=^ 



sin s sin ( s — 5) 
sin a sine 



, „ /sins sin(s — c) 

COS J C = \ : \ , 

\ sm a sin o 



[55] 



tanM-^p 



sins sin(s — a) 



tan 112 



/sin(s — a)sin(s — c) 
\ sins sin (s — b) 



tan*C= A / sin < s - a)si " (s - 6) 
\ sins sin (s — c) 



[56] 



SPHERICAL TRIANGLES. 57 

The foregoing formulae, express the sine, cosine, and tangent of one 
half an angle of a spherical triangle in terms of sides. 

The radical quantities in each one of the formulae must be taken 
positively, because \ A, \B, and \G are each less than 90°, and 
therefore their trigonometrical functions are positive. 

86. The sine, cosine, and tangent of a half side in terms of the 

angles of a spherical triangle. 

Prom the first of [48] we have 

cos A + cos B cos G 



cosa = 



sin B sin G 



• C\\ l--pnqq — 1 cos ^4 + cos ,5 cos C _ cos A -j- cos (B + G) . 

sin B sin G sin B sin C 

/0 \ 1 , -. . cos A + cos B cos C cos A + gos(B— G) 

(Z) l-f-cosa = lH ; : — — = ! * L \ 

sin B sin G sin B sin G 



,.(3) s - m i a _™HA + B + C)cosi(B+C-A) 
\ sin B sin G 



(4 ) cos^ == J cos ^-^ +C ^ os ^^ + ^-^). 
v } 2 \ sin^sinC 

Now, let A + B+C=2S; then B + G - A = 2(S - A); 

A-B + C=2(S-B); and A + B- G=2(S - G); 

and substituting in (3) and (4), they become 



(5) siniq = J- cos ^ cos ^-^ 
V ' 2 \ sin^sinC 



(6) C os±a=J GOS{S ~ B)GOS{S ~ C) ' 
W 2 \ sin.BsinO 

Divide (5) by (6) and obtain 



(7) tani«,=J **S«*(S'-A) 



cos(JS-B)cos(S-C) 

The functions of \o and of -j-c may be obtained from (5), (6), 
and (7), by a simple change of letters. 



58 SPHERICAL TRIGONOMETRY. 

The three systems are as follows : 

4 



cos S cos(S — A) 

Binja = A . p v . n } , 

sinjBsmC 






cosff cos(£ — B) 
sin^lsinC 



sin 



cos -A 



a =V 



cos S cos (S — C) 
sinA sinB 

cos(S-B)eos(S-C) 
sin B sin C 



008*6=^ '-■ 



sin .4 sinO 



cos-Jc 



-v 



qos(S-A)cos(S-B) 
si\\A sin B 



[57] 



[58] 



tania = ^ — 

2 \ cos(£- 



cos(£-^) 



cos (S-.B) cos (#-<?)' 

, " i , / zo'sScosiS — B) 

tmib= \- cos { S-A)L { S-Cy 

tanjc=J- co8^cos(^-0) , 
2 \ cos(£-^L)cos,(#-J3) 

Remark I. The positive sign must be given to the radicals 
of [57], [58], and [59], because -J- a, -J- 6, and -Jc are, each, less 
than 90°. 

Remark II. The functions of J a, J b, and -Jc, in [57] and [59], 
are real quantities. For since the sum of the angles of a spherical 
triangle is greater than 180°,. and less than six right angles, then 
S, or %{A + B + 0), in [57] and [59], is greater than 90°, and less 
than three right angles. Therefore the cos S is either in the second 
or third quadrant, and is negative. The quantities under the radical 
sign are, therefore, positive. 

It is easily shown that cos(# — A), cos(S — B), and cos(# — C) 
are all positive and do not change the result. 



SPHERICAL TRIANGLES. 59 



87. Formula of Delambre. 

From formulas [54] and [55^, we obtain 

(i) 



sine \ sin a sin 6 sine 



(2) 



cos^sin|.B= siB («-«) . >*" sin( S - e) = sin( S -a) cp fi 

sine- \ sin a sin 5 sine 



(3) 



C08i^C08iB = e2» sm(a-a) gin(«-6) = sins gi fi 

sin c \ sin a sin 6 sin c 

(4) 

■inLl sinii? = sin (*- c ) /rin(«-a)»i°(»-6) = sin( 5 -c) sin i a 
sin c \ sin a sin & sin e 

From (1) and (2) by addition and subtraction, we obtain 

/kx sin j-^ cos j- B ± cos-^^4 sin j-J3 _ sin(s — 6) ± sin(s — a) m 
cos-J-0 sine 

and from (3) and (4), we obtain 

•«v cos|-^t cos^-JB if sin^- A sin ^-2? _ sins ~p sin (s—c) 
sin^-C sine 

By Art. 52, 

(7) sin (s — b) + sin(s — a) = 2 sin Jc cos i(a — .&), 

(8) sin(s — &) — sin(s — a)=2cos|-csini(a — b), 

(9) sins -}- sin(s — c) = 2cos-§-c sin -J- (a + b), 
(10) sins — sin(s — c) = 2siniccosi(a -f b), 

also sin c = 2 sin J c cos J c. 

.-. substituting in (5) and (6) the values found in (7), (8), (9), 
(10), and that of sine, we have 



[60] 



60 SPHERICAL TRIGONOMETRY. 

,-j, sin j-( A + B) _ cosj(a — b) 
cos -JO cos-Jc 

^ sin^(^l — .5) = sinj-(g -6) 
cos j-0 sin|-c 

,ov cos^(^4 + .B) _ cosj-(a + 5) 
sin-J-0 cos-^-c 

(4) gosjX^-j) _ sini(g + 5) | 
sin J (7 "" sin-Jc 

Remark. These formulae were discovered by Delambre, in 
1807, and published in Connaissance des Temps for 1809 (p. 443). 
Gauss, to whom they are sometimes attributed, did not publish 
them until two years later, in his work, Theoria motus corporum 

coelestium. 



88. Napier's Analogies. 

If (1) of [60] be divided by (3); (2) by (4); (4) by (3); and 
(2) by (1), we shall obtain the formulae that are known as Napier's 
Analogies. 

They are as follows : 

(1) tanl(^ + B) = coU (7 C0S t (a ~ 5 ? , 
v J 2V ' 2 cos£(a + 6) 



(2) tan-|( J. - S) = cot i C 



sin-|(a — b) 
sin-J-(a + b) 



(3) tan£(a + 6) =tanAc C0S 2(^--B) 



(4) tan J (a 



b) =tanjc 



smj(A-B) 
sin^A + B) 



[61] 



SPHERICAL EIGHT TRIANGLES. 61 



CHAPTER VIII. 

SOLUTION OF SPHERICAL RIGHT TRIANGLES. 

89. The formulae for the solution of spherical right triangles 
are those of [53], Art. 84. 

From formula (1), cos a = cos b cose; it follows that either all 
the cosines are positive, or but one is positive. 

Therefore, in a right triangle, either all the sides are less than 
quadrants, or one side is less than a quadrant, and the other two sides 
are greater than quadrants. 

From (5) and (7) of [53], it follows that tan b and tan B have 
the same sign, and are, therefore, either both greater than 90°, or 
both less than 90°. When such is the case, b and B are of the same 
species. The same is true of c and C. When one is greater than 90°, 
and the other less than 90°, they are then of different species. 

90. The ten equations of [53] constitute but six cases : 

1. Given the two sides ; to find the other parts. 

2. Given a side and its opposite angle ; to find the other parts. 

3. Given the two angles ; to find the other parts. 

4. Given the hypothenuse and one side ; to find the other parts. 

5. Given the hypothenuse and an angle ; to find the other parts. 

6. Given a side and an adjacent angle ; to find the other parts. 

91. Napier's Circular Parts. Two rules, called Napier's Rides 
of Circular Parts of a spherical triangle, include all the possible 
cases. 

The circular parts of a spherical triangle are five: the two sides 
6 and c, the complements of the angles B and C, and the complement of 
the hypothenuse a. The complements of B, C, and a are generally 
written co. B, co. C, and co. a. The right angle is always excluded. 

An examination of Fig. 28 shows that if any three parts be 
taken, as b, c, and co. B, all are adjacent ; but if b, c, and co. a be 
taken, then one is separated from the other two by the remaining 



62 



SPHERICAL TRIGONOMETRY. 



parts co. B and co. C No other arrangements of the five parts, 
when taken by threes, can be made. The three are either adjacent, 
or one is separated from the other two by intervening parts. 

When the parts are adjacent, as co. C, co. a, and co. B, then one 
is called the middle part, and the other two the adjacent parts. 

When the parts are separated, as co. a, co. B, and b, one, b, is 
the middle part, and the other two, co. a and co. B, are called the 
opposite parts. 

co.B t 




co.C 




Fig. 28. 



Fig. 29. 



Let Fig. 29 be a spherical right triangle, in which b and c are 
taken as before ; but the angle B is represented by co. B, or 90° — B, 
the angle C by co. C, or 90°— C, and the hypothenuse a by co. a, or 
90°- a. 

A comparison of Figs. 28 and 29 will show that Napier's Rules 
apply in either case. 

92. Napier's Rules. 

1. The sine of the middle part equals the product of the tangents 
of the adjacent parts. 

2. The sine of the middle part equals the product of the cosines 
of the opposite parts. 



93. That Napier's Rules, the formulas of [53], Art. 84, and the 
six cases of Art. 90, all agree, we shall now show. 

Formula (8) of [53] is 

cos a = cot B cot C. 



SPHERICAL RIGHT TRIANGLES. 



63 



By Napier's first rule, 

sin (co. a) = tan (co. B) tan (co. C). 

Replacing co. a, co. B, and co. C by their values, the result 

becomes 

sin (90°_ a) = tan (90°- B) tan (90°- C), 

or cos a = cot B cot 0. 

Again, formula (9) of [53] is 

cosi?='cos& sin C. 

By Napier's second rule, 

sin (co. B) = cos b cos (co. (7), 

or sin (90°-5) = cos b cos (90°- C), 

or cos 5 = cos b sin (7. 

Six cases are to be considered. 



94. Given b and c ; to find the other parts. 
From (5), [53 J, tan b — sine tan B; .: tanJB = 

(7), tanc = sin 6 tanO; .*. tan (7 = 



tan b 
sine 
tanc 
sin b 



(i), 



cos a = cos b cose. 



It is clear that no ambiguity exists in the solutions, since not 
one of the quantities required is to be found from its sine. 



95. Given b and B ; to find the other parts. 
From 
(2), [53], sin b = sin a sin B-, .-. sina = 

(9), cos B = cos b sin (7; .*. sin (7 = 

(5), tan& = sine tan JB; 



sine 



sin b 
sini? 
cos B 
cosb 
tan b 
tan 5* 



In this case there is nothing to decide whether 
a, C, and c should be greater or less than 90°; there- 
fore the solution remains ambiguous, as the parts are 
determined from their sines. It is the only ambiguous 
case that occurs in spherical right triangles. 

Let ABC (Fig. 30), right-angled at A, be a triangle 




64 SPHERICAL TRIGONOMETRY. 

satisfying the given conditions. Prolong BA and BO until they 
meet at B'. Then the right triangle AB'C also satisfies the given 
conditions ; for it has the angle B' equal to B, the angle A a right 
angle, and the side AC the same as in ABC. 

96. Given B and O ; to find the other parts. 
From (8), [53], cos a = cot B cot C: 

(9), cosJ3 = cos6sin(7; .\ cos 6 



(10), cos C = cos c sin B ; .-. cose 



sinC 
cos C 



smB 
There is no ambiguity in this case. 



97. Given hypothenuse a and a side b ; to find the other parts. 

cos a 

From (1), [53], cos a = cos b cose; .-. cosc= 

cos b 

(2), sin b =sina sinJB; .-. sin 5 = 



(4), tan b = tan a cos C ; .-. cos(7 = 



sin a 
tan b 



tan a 

There can be no ambiguity in this case, except in the value of B, 
but this is removed by the consideration that B and b are always of 
the same species. 

98. Given hypothenuse a and an angle C; to find the other parts. 

From (8), [53], cos a = cot B cot C; .-. cot B = ^~ 

cotj o 

(4) , tan b = tan a cos (7. 

(3), sin c = sin a sin (7. 

There is no ambiguity in (3), when it is remembered that c and 
O are of the same species. 

99. Given a side b and an adjacent angle C; to find the other 
parts. 

From (9), [53], cos B = cos b sin O. 

(7) , tan c = sin b tan C. 

(4), tan b = tan a cos C; .-. tan a = an • 

cos C 

There is no ambiguity in this case. 



SPHERICAL RIGHT TRIANGLES. " 65 

100. Isosceles and Quadrantal Triangles. 

An isosceles triangle is readily solved by dividing it into two 
right triangles by a perpendicular from the angle included by the 
equal sides, then applying formulae [53]. 

A quadrantal triangle which is the polar triangle of a right 
triangle, has one of its sides equal to 90°. When such a triangle 
appears, it is easily solved by means of its polar triangle. 

101. Examples. 

SET VII. 

Isosceles Triangles. 

1. In a spherical right triangle, c = 145°, and A = 23° 28', 
C being the right angle. Find the other parts. 

2. Given c = 32° 34', and A = 44° 44'. Find B, a, and b. 

3. Given a =141° 11', and c =127° 12'. Find A, B, and b. 

4. Given a = 35° 44', and A = 37° 28'. Find B, b, and c. 
This problem gives two triangles. 

5. Given a =118° 54', and B = 12° 19'. Find A, b, and c. 

6. Given A= 91° 11', and 5=111° 11'. Find a, b, and c. 

7. Given a = 1°, and 5=100°. Find A, 5, and c. 

8. Given A = 23° 28', and b = 49° 17'. Find B, a, and c. 

9. Given a = 37° 48', and c = 66° 32'. Find A, B, and 6. 

10. Given a = 59° 38' 27", and b = 48° 24' 16". 
Find A, B, and c. 

11. Given 5=111° 14' 37", and b =121° 26' 25". 
Find ^4, a, and c. 

Quadrantal Triangles. 

12. Given c=90°, A= 54° 43', and B = 42° 12'. 
Find the other parts. 

13. Given c = 90°, A =112° 2' 9", and 6 = 67° 3' 14". 
Find the other parts. 

14. Given c = 90°, a= 22° 53' 30", and 6 = 51° 4' 35". 
Find the other parts. 

15. Given the quadrantal side AB, the angles A and B ; to find 
the other parts. 



66 * SPHERICAL TRIGONOMETRY. 

16. Prove sin 2 -J- c = sin 2 £ a cos 2 \ b + cos 2 \ a sin 2 J b. C being 90°. 

17. Prove tan 2 -J 6 = tan £ (c + a) tan \ (c — a). 

18. Prove tan 2 -J- J. = sin (c — b) cosec(c + 6). 

19. Prove tan 2 ^c = - cos (A +B) sec(A -B). 

20. Prove tan 2 (45°- %b) = sin (J. - a) cosec(^H-a). 

21. In a right triangle show that 

sin (c — 6) = 2sin 2 |-J.cos&sinc. 

22. C = 90°. Show that sin a cos 6 = tan -J ^1 sin (6 + c) . 

23. The equal sides of an isosceles triangle are, each, 45°, and 
the angle included is 95°. Find the other parts. h 

24. In a right triangle, if be the length of the arc drawn from 
C, perpendicular to the hypothenuse c, show that 



cot# = Vcot 2 a + cot 2 6. 

25. a is one side of an equilateral triangle. Find the angle A. 

26. Show that ^ = !illM. 

cos b sin2jB 

27. Required the diedral angles made by the faces of the regular 
polyedrons. 

28. Show that sin (c — a) = sin b cos a tan \ B. 

29. If A = 36°, 5 = 60°, and (7=90°, show that 

a + b + c = 90°. 

30. O being a right angle, show that 

sin J.sin26 = sincsin2jB. 



SPHERICAL OBLIQUE TRIANGLES. 67 



CHAPTER IX. 

SOLUTION OF SPHERICAL OBLIQUE TRIANGLES. 

102. We shall consider only those triangles each of whose parts 
is less than 180°, keeping in mind always the following principles : 

(1) The greater side is opposite the greater angle, and conversely, 

(2) Each side is less than the sum of the other tivo. 

(3) The sum a-\-b -j-c is less than 360°. 

(4) The sum A+B + is greater than 180°. 

(5) If A+B + C >1$0°, then A>1$0°-B-C, 

B>18Q°-A-C, 
C>180°-A-B. 

(6) A side differing more from 90° than another side is in the 
same quadrant as its opposite angle. 

(7) An angle differing more from 90° than another angle is in the 
same quadrant as its opposite side. 

(8) Two sides at least are in the same quadrants as their opposite 
angles respectively. 

(9) The sum of two sides is >, =, or < 180°, according as the 
sum of the two opposite angles is >, =, or < 180°. 

103. Case I. Given the three sides; to find the angles. 

Formulae [54], [55], and [p$], which express the sine, cosine, 
and tangent of one-half an angle in terms of the sides of a spherical 
triangle, may be used. The preference is to be given to [p$]. 

Example. Given the three sides, 

a = 113° 2'56".64, 
6= 82° 39' 28".40, 
c= 74° 54' 31".06 ; 

to find the angles A, B, and C. 



68 



SPHERICAL TRIGONOMETRY. 



Solution : 

a = i(a + b + c) = 135° 18' 28".05 
s-a = 22°15'31".41 

8-6 = 52°38'59".65 

s-c = 60°23'56".99 



log sins =1.847139 

log sin (s- a) = 1.578398 
log sin (s- b) = 1.900336 
log sin (s-c) =1.939264. 



(1) tanj^ = JgHL( s - & ) sin ( g - c ). 
\ sin s sin (s — a) 



(2) taniJ5 = ^ 



logsin(s- 6) = 1.900336 

log sin (s-c) =1.939264 * 

cologsins =0.152861 

colog sin(s - a) = 0.421602 

2) 0.414063 

log tan 1 J. = 0.207031 

\A= 58°10'1".10. 
.-. A = 116° 20' 2".20. 



sin s sin (s — 6) 

log sin(s- a) = 1.578398 

log sin (s-c) =1.939264 

cologsins =0.152861 

colog sin(s - b) = 0.099664 

2)1.770187 

log tan i£ = 1.885093 

i£ = 37°30'25".8. 
.-. J5 = 75° 0'51".6. 



(3) tan|(7 

log sin 1 

log sin (s — 6) 



/sin(s — a) sin(s — b) 



sin s sin (s — c) 
log sin (s-a) = 1.578398 



1.900336 

cologsins =0.152861 

colog sin (s-c) = 0.060736 

2)1.692331 

log tani C= 1.846166 

i C = 35° 3' 29".58 
.-. C = 70° 6' 59".16. 



104. Case II. Given the three angles; to find the sides. 

Formulae [57], [58], and [59] are applicable. But the three 
formulae of [59] are generally to be preferred. 



SPHERICAL OBLIQUE TRIANGLES. 69 

105. Case III. Given two sides and the included angle; to find 
the other parts. 

We obtain the simplest solution of this problem by means of 
Napier's Analogies, formulae [61], Art. 88. 

If the given parts be a, b, and C, we compute at once A and B 
by (1) and (2) of [61] ; then c by (3) or (4). 

Example. Given a = 113° 2' 56 ".64, 
&=' 82° 39' 28".40, 



C = 138° 50' 13".69 ; to find A, B, and c. 



Solution : 



i ( a - 6) = 15° 11' 44".12 log sin £ (a - b) = 1.418492 

\ (a + o) = 97° 51' 12".52 log cos | (a - b) = 1.984544 

%C =69° 25' 6".845 log sin £ (a + 6) = 1.995907 

logcos£(a -f &) = 1.135578 
log cot i C =1.574616 

(1) tan j- (A +B) = cot|(7 cos ^ (a ~ b) > 

(2) tan i (A - B) = cot i C sm * (a - b "> - 
W 2V ; 2 sini(a + &) 

logcot^-C =1.574616 log cot i C =1.574616 

logcos£(a-6) =1.984544 log sin J (a - b) =1.418492 

cologcosi(a + 5) =0.864421 cologsin£(a -f- b) =0.004093 

log tzni(A+B) = 0.423581 log tan J (4-5) = 2.997201 

.-. i (A+B) = 110° 39' 35".29. .-. } (A-B) = 5° 40' 26".91. 

i(A+B) = 110°39'35".29, 
i(A-B)= 5°40'26".91. 

.-. A = 116° 20' 2".20, 
B = 104° 59' 8".38. 

log cosi (A +£) = 1.547551. log cosi (A -B) = 1.997867. 



70 SPHEEICAL TEIGONOMETRY. 

(3) tan^c = tan j- (a + b) G0S H A +j£ . 

. , w .. j log sin* (a + 6) =1.995907 
logtan4-(a-f &) = < , 

v y (colog cos£ (a + 6) = 0.864421 

log cos i (^4+JB) = 1.547551 

colog cosi (A-B) = 0.002133 

log tan|c = 0.410012 

jc= 68° 44' 32".30. 
.-. c = 137°29' 4".6. 

106. Case IY. Two angles and the included side given ; to find 
the other parts. 

If the given parts be A, B, and c, we compute a and b by 
formulae (3) and (4), and C by (1) or (2), of Napier's Analogies. 

107. Case V. Given two sides and an angle opposite one of 
them; to find the other 



If the given parts be a, b, and A, we compute B at once by the 

formula m = m , and then obtain the corresponding value of 
sin b sin a 

C by (1) or (2) of Napier's Analogies, and the value of c by (3) 
or (4). 

In order that the problem may be possible, it is necessary that 
the sinJ5 or sin& be comprised between zero and 1. When this 
condition is satisfied, then B or b will have two values, supplemen- 
tary to each other. But it is necessary that the corresponding 
values of tan -J- C and tan^-c be positive, which requires that A—B 
and a — b have the same sign. If this Condition be not satisfied 
for either of the two values of B or 6, the problem admits of no 
solution. But, if it is satisfied, a solution will necessarily follow, 
A—B and a — b being of the same sign, C and c will be comprised 
between zero and 180°, by means of two formulas of Napier. These 
values of C and c will be the same as those which the other two 
formulas of Napier will give. 

Example. Given a = 50° 45' 20", 
b = 69° 12' 40", 
A = 44° 22' 10" ; to find B, C, and c. 



Solution 



SPHERICAL OBLIQUE TRIANGLES. 71 

sin A sin b 



(1) sin 5 = 



sin a 



log sin 69° 12' 40" = 1.970763 

log sin 44° 22' 10" = 1.844652 

colog sin 50° 45' 20" = 0.111004 



log sin£ = 1.926419 

.-. B = 57° 34' 51" A, 

or 122° 25' 8".6. 
There are then two solutions. 

(2) cotj-C = cos *(* 'l± a ] ttmi(A+B). (1) of [61]. 

COS 2" (0 — ft J 

i {B +A) = 50° 58' 30".7 log cos 59° 59' = 1.699189 

i (B-A)= 6°36'20".7 log tan 50° 58' 30".7 =0.091246 
|(6-f a) = 59° 59' colog cos 9° 13' 40" = 0.005657 

J(6-o)= 9° 13' 40" log cot |(7 = 1.796082 

i(7= 57°58'55".3. 

When B = 57° 34' 51 ".4, then (7 = 115° 57' 50".6. 

But when £=122° 25' 8".6, 

then -§-(£-M) = 83°23'39".3, 

i(B-A)= 39° 1'29".3, 

J (6 + a) = 59° 59', 
J (6 - a) = 9° 13' 40". 

By (1) of [61], 

log cos 59° 59' =1.699189 

log tan 83° 23' 39".3 = 0.936270 

colog cos 9° 13' 40" =0.005657 



log cot i(7 = 0.641116 
.-. i C = 12° 52' 15".8 
For second value of B, then (7 = 25° 44' 31 ".6 



72 SPHERICAL TRIGONOMETRY. 

(3) The values of c are found by (3) of [61]. 

2 cos^(B-A) 2K T J 

log cos 83° 23' 39".3 = 1.060837 

logtan59°59' =0.238269 

colog cos 39° V 29".3 = 0.109650 



log tan fc = 1.408756 

£.c = 14° 22' 32".6. 

.-. c = 28°45' 5".2. 

The other value of c is found by taking the other values of 
i(B+A) and i(B-A). 

108. Case VI. Given two angles and the side opposite one of 
them; to find the other parts. 

This case gives rise to the same ambiguities that are found in 
Case V. 

To solve the problem, the formulae to be used are 

/iX • , sin a sin B 

(1) sin6 = : — - — ; 

sini 

(2) Napier's Analogies, [61]. 

109. The two cases that admit of two solutions. 

Case V. and Case VI. are the only two cases of spherical 
triangles that admit of two solutions, and yet they do not always 
admit of such ambiguity. The formulae which relate to these two 
cases make known the number of solutions, and determine without 
ambiguity the elements of each of them. 

In order that the problem may be possible, it is necessary and 
sufficient that tan^L and cos a, cos A and tana, have the same 
sign ; that is to say, that A and a be both less than 90°, or both 
greater than 90°. There is then but one solution. 

Passing to the general case, it is necessary, in order that the 

problem be possible, that s sm be less than 1 ; if this condition 

sin a 

be satisfied, there are then two values of B that satisfy the equation 

sin£ = sm&sin ^ one of which is B and the other 180° -B. 
sin a 



SPHERICAL OBLIQUE TRIANGLES. 73 

Now, A—B and a — b must have the same sign, in order that 
B and 180°— B may satisfy the problem. 

In the cases in which two solutions are indicated, no solution is 
possible if sin a be less than sin b sin A 

If a lies between b and 180 — b, there will be one solution. 

If a does not lie between b and 180° — b, either there are two 
solutions or no solution. The cases in which a = b, or a = 180°— b, 
are not included in the last supposition. 

110. Examples. 

SET VIII. 

1. Given a = 43° 27' 36", 

b= 82° 58' 17", 
A = 29° 32' 29" ; to find B, C, and c. 

In this problem, A < 90°, and b < 90° ; therefore a < b will give 

two solutions. 

Take smB = smbsinA . 

sin a 

2. Given a = 74° 23', 

b= 35° 46' 14", 

c = 100° 39" ; to find A, B, and C. 

3. Given A = 48° 30 r , 4. Given a = 70° 14' 20", 

B = 125° 20', b = 49° 24' 10", 

0= 62° 54'; c = 38° 46' 10"; 

to find a, b, and c. to find A, B, and (7. 

5. Given A = 129° 5' 28", 6. Given a = 68° 46' 2", 
5 = 142° 12' 42", b = 37° 10', 

(7 = 105° 8' 10"; = 39° 23' 23"; 

to find a, 6, and c. to find A, B, and c. 



7. Given .4 = 34° 15' 3", 8. Given a = 97° 35 



B= 42° 15' 13", 6 = 27° 8' 22" 

c= 76° 35' 36"; A = 40° 51' 18" 

to find a, 5, and C. to find 2?, (7, and c. 



74 



SPHERICAL TRIGONOMETRY. 



9. Given A = 50° 12', 
B = 58° 8', 
a= 62° 42'; 
to find 6, c, and (7. 
How many solutions has (9)? 

11. Given A = 50°, 
a= 40°, 
6= 60°; 
to find B, G, and c. 

13. Given A = 30° 28' 11", 
JB = 130° 3' 11", 
c= 40°; 
to find a, b, and 0. 

15. Given A = 31° 34' 26", 

B= 30° 28' 12", 
c= 70° 2' 3"; 
to find a, 5, and C. 

17. Given ^L = 32° 26' 6".66, 
.B == 130° 5' 22", 
a= 44° 13' 42"; 
to find b, c, and G. 



10. Given a = 150° 17' 23", 
6= 43° 12', 
c= 82° 50' 12"; 
to find A, B, and 0. 



12. Given ^ = 135° 5'28".8, 
C = 50° 30' 8".4, 
b= 69°34'55".9; 
to find a, c, and 5. 

14. Given a= 68° 46' 2", 
6= 43° 37' 38", 
c= 37° 10'; 
to find .4, B, and (7. 

16. Given a= 63° 50', 
6= 80° 19', 
a-"=F 51° 30'; 
to find B, C, and c. 

18. Given J. = 120° 43' 37", 
B = 109° 55' 42", 
G = 116° 38' 33" ; 
to find a, b, and c. 



AKEA OF SPHERICAL TRIANGLES. 



75 



CHAPTER X. 



AREA OF SPHERICAL TRIANGLES. 

111. Given the angles of a spherical triangle; to find its area. 

Let ABC (Fig. 31) be a spherical triangle traced upon the 
surface of a sphere, and CAE, CBE, AED, ABD, and ACD be 
semicircumferences. Then the triangle ABC will be a part of each 
of three lunes, CAEB, CAB-BED, and ABDC. 




Fig. 31. 

The surface of the hemisphere whose base is ACDE, is equal 
to the surface of the three lunes less twice the triangle ABC. Let 
r = the radius of the sphere. 

(1) The surface of the hemisphere = 2irr 2 , 

(2) The surface of the lune CAEB = 2 Cr 2 , 

(3) The surface of the lune CAB -BED = 2 Br 2 , 

(4) The surface of the lune ABDC = 2Ar. 



76 SPHERICAL TRIGONOMETRY. 

Therefore 2 ABC = 2 (A + B + C - 7r)r 2 ; 

or ABG = (A + B+C-Tr)r>. 

Denote A + B 4- C — ir, which is called the spherical excess by 
E, the value of ABC then becomes Er 2 . 

But Er 2 = ~ x2in*, 

2tt 

•'• AB0 = m? * 2 ^ = # * "* [62] 

112. Given £/ie £/iree sides o/ a spherical triangle; to find its 
area. 

An elegant formula due to Simon l'Huillier, of Geneva, furnishes 
a direct method for the solution of this problem. 

Since E = A + B+ C - tt, 

then t^iE = s ^^±^±^4 

4 COSi(>l + £ + C-7r) 

= sin j (A + B) - sin | ( C - tt) 
cos ^ (A -f- 5) + cos -J- (0 — tt) 

= sin j(A + £) - cos j- O . 
cos|(^ + J5) + sin£<7' 

multiplying both terms of this fraction by cos-Jc, it becomes 

sinj(A + B) cos-|-c — cos \ c cos j- C 
cos|-(^l -f- B) cos-Jc + cos-J-c sin-J-C 

which by (1) and (3) of [60], 

_ [cosj-(q — b) — cos -|c] cos j-0 
[cos£(a + b) + cos Jc] sin -J- C 

and replacing cos-J-0 and sin -JO by their values, [55] and [54], we 
have this last 



__ cos -j- (a — b) — cos \ c j sins sin (s — c) 
cos|-(a + b) + cos^-c \ sin(s — a) sin(s — 6)' 



AREA OF SPHERICAL TRIANGLES. 77 

which, by PL Tr., 



_ sin \{a — 5 + c) sin \(b -\-c — a) I sins sin (s — c) 
cosi(a + 6 + c) cos£(a-f & — c) \sin(s — a) sin (5 — b)' 

and since s = ^(a + 6-f-c), 

this last fraction becomes 



/ sin 2 ^(s — b) sin 2 ^-(g — q)sinssin(s — c) . 
\cos 2 -2-s cos 2 -j-(s — c) sin(s — a) sin(s — b) ' 

and since sin s = 2 sin -§■ s cos -J- s, 

sin(s — c) = 2sin-J(s — c) cos-J(s — c), 
sin(s — b) = 2sin-J(s — 6) cos-J(s — 6), 

and sin(s — a) = 2sin|-(s — a) cos^(s — a), 

the foregoing radical then 

/ sin 2 ^-(s — b) sin* -J- (a— a) sin-^-s cos^-s sin |-(s—c) cos ^(s — c) 
\cos 2 -Js cos 2 -|-(s— c) sin-J(s — a) cos-J(s — a) sinj(s— b) cos^-(s— b) 

therefore 



tsai^E = Vtan-J-s tan-J-(s — a) tan-J(s — b) tan-J-(s — c). [63] 

If E in [62] be replaced by its value in [63], then the area 
of a spherical triangle may be easily obtained in terms of its 
three sides. 

113. To find the area of a spherical polygon in terms of its. 
angles. 

Draw arcs of great circles from one of the vertices to the others ; 
the polygon will then be divided into triangles, whose areas may 
be computed separately. Their sum will be the area of the 
polygon. 

If n be the number of sides of a polygon, T the sum of all its 
angles, and P its area, then 

P=[T-(n-2)Try. [64] 



78 



SPHERICAL TRIGONOMETRY. 



114. To find the angular radius of a small circle circumscribed 
about a spherical triangle. 

Let OA, OB, and 00, the arcs of great circles, be drawn from 
0, the pole of the small circle ABC, to the vertices of the spherical 
triangle ABO (Fig. 32). Three isosceles spherical triangles are 
formed, having the sides of the given triangle for their bases. 




Fig. 32. 



Let = each of the equal angles in the isosceles triangle OB A ; 
$', each equal angle in OAO\ and 6", each equal angle in OBG. 
Whether the pole of the circumscribed small circle be taken within 
or without the triangle, the formulae will be the same. 

Taking the pole within, then 

6 + 6'= A; 6 + 6"=B; and 6'+6"=0; 
whence 6 + 0'+ 0"= i(A + B+C) = 90° + ±E. 
Therefore = 90° - ( C - \ E) , 

O' = 90°-(B-±E), 



AREA OF SPHERICAL TRIANGLES. 79 

Let the angular radius 

OA=OB= OC=B. 

Draw OD, an arc of a great circle, from 0, perpendicular to the 
side c of the given triangle. The spherical right triangle ADO 
thus formed gives 

tan B cos = tan \ c. 

Eeplacing 6 by its value 90°— (C — \E), we obtain 

tan-i-c 



tani? = 



cos(90°-(<7-iE)) 



or 



tani2= — *^i^ 



sin(C-iJE) 



[65] 



115. To find the angular radius of a small circle inscribed in a 
spherical triangle. 

Join the pole 0, of the small 
circle EFD (Fig. 33), to the 
vertices, by the arcs of great 
circles OA, OB, and 00. Draw 
OF, OD, and OE, arcs of great 
circles, from 0, perpendicular to 
the three sides of the triangle. 
Let OF= r, AOF being a right 
triangle, one of whose sides is 
r. 
s 



another is \(b + c — a) or 
- a, and the angle OAF ad- 
jacent to s — a equals \A 
Then 



iO. 



tanr = sin(s — a) tani0, 

or, replacing tan 10 by its value 

in terms of the sides of ABC, we obtain 




Fig. 



tanr 



-v 



sin( s — a) sin(s — b) sin(s — c) 
sins 



[66] 



80 



SPHERICAL TRIGONOMETRY. 



116. To find the angular radii of the small circles escribed upon 
the sides of a spherical triangle. 

An escribed circle is one which is tangent to one side of a triangle 
and the other two sides produced. 

Prolong the sides b and c (Fig. 34) of the 
triangle ABC, until they meet at D, 180° from 
A. In this manner, another triangle BBC will 
be formed whose sides are 

(1) a, 

(2) p-6, 

(3) 7T -c. 
E 

Let r', an arc of a great circle, be the radius 
required. From the right triangle DOE, we 
C obtain, by letting A or its equal D — 6, 

tanr' = tan- sins. 

2 

Remembering that s = \ (a + b -f- c), and 
that tan -1-0 is equal to 




Vsin(, 



s — b) sm(s — c) 
sin s sin {s — a) 



then tan r- 



V sin(s — 6) sin(s — c) _ / sing sin (3 — b) sin(s — c) 
sin s sin (s — a) \ 



sm(s — a) 



tanr" 



Vsin( 
- 
s 



a) sin(s — c) _ kins sii(s — a) sin(s — c) 
Af sin(s — 6) 



sins sin(s — b) 



tan' 



g . /sjn(s-a^m(s-&) = J 
\ sins sin (s — c") \ 



sins siu(s — a) sin (5 — b) 
sin(s — c) 



K67] 



r" and r'" are the angular radii of the two small circles tangent 
to the sides b and c and the prolongations of the others. 



AREA OF SPHERICAL TRIANGLES. 81 

117. Examples. 

SET IX. 

1. Find the area of a spherical triangle, whose angles are 140°, 
92°, and 68°, respectively, described on a sphere whose radius is 
15 feet, 

2. What is the area of a spherical triangle whose angles are 
150°, 110°, and 60°, respectively, described on a sphere whose radius 
is 10 feet ? 

3. Given a = 98°, b =110°, and c = 115°; to find the area of the 
triangle when traced on a sphere whose radius is 100 feet. 

4. Each side of a spherical triangle is 10°. Eequired the 
spherical excess, the sphere's diameter being 10 feet. 

5. Given a = 88° 12' 20", b = 124° V 17", and C=50°2'l"; to 
find the spherical excess. 

6. If the angles of a spherical triangle be together equal to four 
right angles, show that 

cos 2 ia + cos 2 i6 -f- cos 2 ic = 1. 

7. A spherical polygon of five sides has A = 75°, B — 80°, 
0= 115°, D= 120°, and E = 150°. Find its area. 



82 SPHERICAL TRIGONOMETRY. 



CHAPTER XI. 

APPLICATIONS OF SPHERICAL TRIGONOMETRY. 

118. The theory of Spherical Trigonometry has one of its most 
useful applications in the solution of astronomical problems. The 
theory owes its origin in no small degree to the inquiries that have 
grown out of the subject of Astronomy. Nearly all the lines 
that are considered as traced on the surface of the earth are, by 
astronomers, extended to the heavens, thus constituting a sphere 
called the Celestial Sphere, whose radius is "greater than any 
assignable quantity." 

119. Definitions of Terms. 

(1) The Horizon is a great circle traced on the celestial sphere, 
whose poles are called the Zenith and the Nadir. The plane of the 
horizon touches the surface of the earth at the point of observation, 
and is then called the Sensible Horizon. But when that plane passes 
through the centre of the earth parallel to the sensible horizon, it 
is called the Rational Horizon. 

(2) The Zenith is a point in the celestial sphere vertically 
overhead. 

(3) The Nadir is a point in the celestial sphere directly opposite 
to the zenith. 

(4) Vertical Circles are great circles passing through the zenith 
and nadir, perpendicular to the horizon. Upon them the altitudes 
of celestial objects are measured. 

(5) The Meridian is the great circle that passes through both 
the North Pole and the South Pole, and the Zenith and the Nadir. 

(6) The Prime Vertical is the vertical circle cutting the meridian 
at right angles at the zenith, and, therefore, having an east and 
west direction on the celestial sphere. 

120. The Equinoctial or Celestial Equator is a great circle traced 
on the celestial sphere by the plane of the earth's equator extended 
to the heavens. 



APPLICATIONS OF SPHERICAL TRIGONOMETRY. 83 

(1) The Axis of the celestial sphere is the axis of the earth 
extended in both directions until it meets the sky. 

(2) The North Pole and the South Pole are the ends of the axis 
of the celestial sphere. 

(3) Hour Circles, sometimes called Celestial Meridians, are great 
circles of the celestial sphere passing through its two poles, and 
perpendicular to the equinoctial. 

121. The Ecliptic is a great circle traced by the sun in its 
apparent annual motion about the earth. The Ecliptic and the 
Equinoctial intersect at an angle of nearly 23° 27', at two points, 
one, called the Vernal Equinox, the other, the Autumnal Equinox. 
The date of the Vernal Equinox, i.e., when the sun crosses the 
equinoctial going north, is the 20th of March ; that of the Autumnal 
Equinox, the 22d of September. 

(1) Circles of Celestial Latitude are great circles that pass through 
the poles of the ecliptic, perpendicular to the plane of the ecliptic. 

(2) The point on the ecliptic from which celestial longitude 
is estimated is the vernal equinox, towards the east. 

122. The position of a star in the celestial sphere may be 
described in several ways by means of great circles and their poles 
taken as standards of reference. Whatever these standards may 
be, the quantities employed are called the Spherical Co-ordinates 
of the star. 

The standards of reference and the spherical co-ordinates are 
as follows : — 

f Azimuth, and 

(1) Horizon and Zenith < Altitude or Zenith 

I Distance. 



(2) Equinoctial and North Pole 



Eight Ascension, and 
Declination ; or, 
Declination, and 
L Hour-Angle. 



, Latitude, and 

(3) Ecliptic and Pole { T ' 

v / L Longitude. 



84 



SPHERICAL TRIGONOMETRY. 



Let ENWS be the plane of the horizon (Fig. 35) ; Z, the zenith ; 
0, the point of observation ; WZE, the prime vertical ; NZS, the 
meridian ; and B, any star. 




Fig. 35. 



(1) The Azimuth of a star is its angular distance from either 
the north or the south point of the horizon to a vertical circle 
passing through the star. 

The Azimuth of B, from the south point S, is the arc AS. 

(2) The Altitude of a star is its angular elevation above the 
horizon, measured on a vertical circle passing through the star. 

The Altitude of B is the arc AB. The Zenith Distance of B is 
BZ. The altitude + the zenith distance = 90°. 

(3) The Right Ascension of a star is the arc of the equinoctial 
included between the vernal equinox and the foot of the hour-circle 
passing through the star. 

Let MWTE (Fig. 36) be the equinoctial ; B, the vernal equinox ; 
P and R, the poles of the equinoctial ; POR, the axis of the celestial 
sphere ; SWNE, the horizon ; and A, any star. 



APPLICATIONS OF SPHERICAL TRIGONOMETRY. 



85 



The Right Ascension of the star A is the angle BPD, or the arc 
BD measured from the vernal equinox B. 

The Hour-Angle of the star A is the angle DPM, or the arc MD. 




Fig. 



(4) The Declination of a star is its angular distance north or 
south of the equinoctial, and is measured by that arc of the hour- 
citcle that extends from the equinoctial to the star. 

The North Polar Distance is the arc of the hour-circle that extends 
from the star to the pole. 

The Declination of the star A is the arc AD, and the North Polar 
Distance is the arc AP. 

Let EBFD (Fig. 37) be the equinoctial ; ABCD, the ecliptic ; 
P, the pole of the equinoctial ; Z, the pole of the ecliptic ; B, the 
vernal equinox ; CBF, the obliquity of the ecliptic ; S, any star ; 
PSK, the hour-circle; and ZSM, the circle of latitude through S. 



86 



SPHERICAL TRIGONOMETRY. 



(1) The Latitude of a star is its angular distance from the star 
to the ecliptic, measured on a circle of latitude. 
The arc SM is the latitude of the star &. 

z 




Fig. 37. 

(2) The Longitude of a star measures the arc between the vernal 
equinox and the point on the ecliptic cut by the circle of latitude 
through the star. 

The arc BM is the longitude of the star 8. 

123. Given the surfs right ascension and declination ; to determine 
his longitude and the obliquity of the ecliptic. 

Let ED RON be the celestial 
meridian, passing through the 
points where the sun's declina- 
tion is greatest; NM, the axis 
of the sphere ; ER, the equator 
or equinoctial; DC, the ecliptic; 
NSM, the declination circle pass- 
ing through the sun S. Then 
APS is a spherical triangle, right- 
angled at P. The right ascen- 
sion is AP; the declination, SP; 




APPLICATIONS OF SPHERICAL TRIGONOMETRY. 



87 



and we are to find the longitude AS, and the obliquity SAP. 
Solve by Napier's Rules for Circular Parts. 



124. Given the sun's declination ; to find the time of 
any place ivhose latitude is known. 



rising at 



Let NEMR (Fig. 39) be the meridian of the place; Z, the 
zenith ; HO, the horizon ; BC, the 
apparent path of the sun on the 
proposed day, cutting the horizon 
in S. EZ, then, will be the lati- 
tude, and EH or OR will be the 
complement of the latitude. EH 
or its equal measures the angle 
OAR. PS is the sun's declina- 
tion, and AP, expressed in time, 
will be the time of sunrise. De- 
grees are changed into hours by 
dividing by 15. Therefore in the 
spherical right triangle APS, we 
have given PS, the declination, 

the angle SAP, to find AP, the time from 6 o'clock. 
Napier's Circular Parts. 




Fig. 39. 



Solve by 



125. Given the obliquity of the ecliptic and the declination of the 
sun ; to find his longitude and right ascension. 

Let OB (Fig. 40) represent 
the ecliptic ; OC, the equinoctial ; 
and P, the pole of the equinoctial. 
Let A be the sun's position, and 
PAD the arc of a great circle 
passing through the sun and the 
pole. Then OAD is a spherical 
triangle, right-angled at D, and 
AD is the sun's declination, AO 
his longitude, and AOD the ob- 
liquity of the ecliptic. AD and 
the angle AOD are given, and it 
is required to determine the longi- 
tude and right ascension. 




Fig. 40. 



SPHERICAL TRIGONOMETRY. 




Fig. 41. 



126. Given the hour-angle, latitude, and declination of any star; 
to determine the altitude and azimuth. 

In Pig. 41, PS = the complement of the dec- 
lination, PZ equals the complement of the lati- 
tude, and the angle SPZ, the hour-angle. The 
angle SZP is the supplement of the azimuth, 
and the arc SZ is the complement of the altitude. 
P is the pole ; Z, the zenith ; and S, the star. 
This problem gives 90°— d, 90° — I, and the hour- 
angle SPZ, from which to determine the altitude 
and the azimuth. 

The angle SZP may be found by (1) and (2) 
of Napier's Analogies ; then by the law of sines 
SZ can readily be obtained. 

127. Examples. 

SET X. 

1. Given the sun's right ascension on a certain day, 53° 38', and 
declination 19° 15' 57"; to determine his longitude and the obliquity 
of the ecliptic. 

2. On a certain day the sun's declination was observed to be 
4° 13' 31J", the obliquity of the ecliptic being 23° 27' 51"; required 
his right ascension. 

3. Given the sun's declination 23° 28' ; what is the time of sun- 
rise at latitude 52° 13' K ? 

4. Eequired the time of sunrise at latitude 57° 2' 54", when 
the sun's declination is 23° 28' N. 

5. The obliquity of the ecliptic on a certain day was 23° 27' 36", 
and the declination of the sun 21° 52' 56". Find the sun's longitude 
and right ascension. 

6. On a certain day the declination of the sun was 5° 58' 8", 
and the obliquity of the ecliptic 23° 27' 38". Find the sun's longi- 
tude and right ascension. 

7. The latitude of a star is 40° 36' 23".9, the declination 
23° 4' 24".3, and the hour-angle 46° 40° 4".5. Find the star's altitude 
and azimuth. 

8. The latitude of a star is 40° 36' 23".9, its altitude 47° 15' 18".3, 
and the azimuth 80° 23' 4".47. Find the star's decimation and hour- 
angle. 



MISCELLANEOUS EXAMPLES. 89 



CHAPTER XII. 

MISCELLANEOUS EXAMPLES. 

128. SET XI. 

1. The sides of a triangle are 17, 21, and 28. Prove that the 
length of a line bisecting the longest side and ' drawn from the 
opposite angle is 13. 

2. Show that the area of a quadrilateral, whose diagonals a and b 
intersect at an angle A, is ^ab sin A. 

3. If + #'+#"= 180°, prove that 

(a) sm20 + sin20'+ sin20" = 4sin0 sinfl'sinfl"; 
(6) sin {0 + $') sin ($' + 0") = sin0 sin0". 

4. If + $' + $" = 90° ; prove that 

(a) cos20-f-cos20'+cos20''=l-f-4sm0sin0'sin0''; 
(6) tan£0 + tan£0'+ tani0'" ; - tan|0 tan£0'tan|-0" 

= 1 - tan-J-0 tan£0' — tan J0 tan£0" — tan£0'tan£0". 

5. If tan 6=i, and tan0'=|, show that 2(9 + (9' =45°. 

6. If cot 2 = - tan 0', show that tan (0 - 0') = cot 0. 

7. If cos30 + cos20 + cos0 = O, show that 6=4,5°, or 120°, or 
135°, etc. 

8. In a triangle ABC, right-angled at C, show that 

sin 2 iM = — ; cos 2 ^=^-^; tan 2 f4 = ^^. 
2c 2c ' c-\-b 

9. In a triangle ABC, right-angled at C, show that 

S in2^ = -^L, and cos2^ = ^l— % 
a 2 + 6 2 ' a 2 + 6 2 

10. In a triangle ABC, right-angled at C, show that the 
area = £c 2 sin2^L = -A-a 2 tan B = -J- 6 2 tan A 



90 PLANE AND SPHERICAL TRIGONOMETRY. 

11. If cosi3sinO = sin^l, show that the triangle will be isos- 
celes. 

12. The triangle ABC has its angles A, B, and C, in the propor- 
tion of the numbers 2, 3, and 4, respectively ; show that 

n A a-\-c -. ,o a(a-\-c) 2 

cos£J.= p. and 5 2 = v ^ y • 

2 26 2a + c 

13. tan" 1 — tan- 1 — — = — : show that x = V3 + 1. 

x — 1 a; + 1 12 

14. If (s — a) (s — 6) = ab, show that the triangle is impossible. 

15. The area of a triangle is 84 square inches, and two of its 
sides are 15 and 13 inches, respectively ; find the third side. 

16. The sides of a triangle are 3, 7, and 8, respectively. Com- 
pare the radii of the inscribed and circumscribed circles. 

17. Two sides of a triangle are 8 and 10 inches, respectively, 
and the included angle 30°. Find the area. 

18. A, B, C, D are four trees in a row, such that AB, BC, CD 
subtend equal angles at a point P. If AB = 40 feet, BC = 20 feet, 
and CD = 60 feet, show that PA== 24 V5 feet, PB = 8 VlO feet, 
PC = 12 V5 feet, and PD = 24 VlO feet. 

19. A lighthouse 60 feet high is just seen from the deck of a 
ship 12 feet above the water ; how far is the ship from the light- 
house ? 

20. If the length of an arc of 60° is 11 feet, show that the radius 
of the circle is 10 feet 6 inches. 

21. The supplement of one angle of a triangle is double the 
complement of another, and triple that of the third ; find the 
angles. 

22. A ship sailing 1ST. sees two lighthouses due E. After sailing 
an hour they are S.E. and S.S.E., the distance between them is 8 
miles ; find the rate of the ship. 

23. Show that the number of acres in a field, whose sides are 
400, 300, 300, and 300 yards, respectively, and one angle adjacent 

to the largest side is a right angle, is — - (5Vll + 24). 



MISCELLANEOUS EXAMPLES. 



91 



24. In Fig. 42, the line ED = 200 yards, DC = 200 yards, and 
EF = 200 yards. The angle AFE = 83°, BDE = 156° 25', BDC = 
54° 30', ^LEJD = 53° 30', AEF= 54° 3V, and J3CD = 88° 30'. The 
length of J.J5 is required. 




Fig. 42 



25. Given .45 = 800 yards, AC = 600 yards, and BC = 400 yards. 
The angle .IPO = 33° 45', and £PC= 22° 30'. Find the distances. 
AP, GP, and BP. (Fig. 43.) 




Fig. 43. 



26. A balloon is observed from two stations 3000 feet apart. 
At the first station the horizontal angle of the balloon and the other 
station is 75° 15', and the elevation of the balloon is 18°. The hori- 
zontal angle of the first station and the balloon, measured at the 
second station, is 64° 30'. Find the height of the balloon. 

27. In a spherical equilateral triangle, show that 

2cos?sin^ = l. 



92 PLANE AND SPHERICAL TRIGONOMETRY. 

28. In a spherical triangle, if 6 + c = ir, show that 

sin2i3 + siii2<7 = 0. 

29. ABC is a spherical right triangle, in which A is not the 
right angle. Show that \lA — a, then c and b are quadrants. 

30. In a spherical triangle, if ^1 = -, .B = —, and C =-, show 

7r 2 5 3 

that a + b -+- c = -• 

31. A spherical square is divided into four equal right triangles 
by two diagonal arcs. Find the angle A of the square and one of 
its sides a. 

32. If c x and c 2 be the two values of the third side of a spherical 
triangle, when A, a, and b, are given, and the triangle is ambiguous, 
show that 

tan-j-C! tan-J-c 2 = tan-J-(6 — a) tan J (b + o). 

33. If the equal sides of a spherical isosceles triangle ABC be 
bisected by an arc DE, and BC be the base, show that 

sin i BE = i sin £ BC sec £40, 

34. The sides of a spherical triangle are 105°, 90°, and 75°, 
respectively ; find the sines of all the angles. 

35. Find the area of a regular spherical polygon, whose angles 
are 50°, 95°, 130°, 140°, and 160°, respectively, on the surface of a 
sphere whose radius is 15 feet. 



TRIGONOMETRIC TABLES. 



93 



CHAPTER XIII. 



TRIGONOMETRIC TABLES. 



129. In order to use the trigonometric functions, it is necessary 
to compute the values of the functions of a given arc, and reciprocally 
be able to find the value of an arc when the value of one of its 
trigonometric functions is known. To reach this end, it is indis- 
pensable to have a table which will make known the values of 
the functions corresponding to the successive values of an arc 
comprised between 0° and ^ir, the intervals being sufficiently small. 

Certain preliminary propositions must first be established, in 
order to show how such a table may be computed. 

130. Theorem I. Every arc comprised between 0° and \-k is 
greater than its sine and less than its tangent. 




Fig. 44. 

Let AD (Fig. 44) = 0, be an arc comprised between 0° and |tt; 
DE, the sine, and AB the tangent of 0. Prolong BE to F, and 
draw the tangent BC; we shall have the arc DAF>DF, and 
ADC<AB + BC. 

The arc is one-half the arc DAF or the arc ADC; sin0 is one- 
half DF, and tan0 is equal to each one of the lines AB and BC; 
therefore 6 > sin 0, and 6 < tan 0. 



94 CONSTRUCTION OF TABLES. 

131. Corollary. If the arc decreases from ^tt to 0, the ratio 

approaches unity as a limit. 

6 

Since tan0 = ^L# 

COS0 

we may write sin < < ^-^, 

COS0 

from which, dividing by sin 0, 



sin0 cos0 
Hence it follows that the ratio — — - is comprised between unity 

^ SH10 

and the fraction , whose limit is unity when = : 

cos 

.-. limit of -^- = 1, or, limit of ^ = 1. 
sin0 ' ' 

132. Theorem II. The excess of an arc comprised between and 
\tv over its sine is less than the one-fourth, and also less than the 
one-sixth, of the cube of that arc. 

A3 

(1) To show that — sin < — , it will suffice to consider the 
inequality established in Theorem I. ; 

tan£0>£0. 

Multiplying by 2 cos 2 ! = 2(1- sin 2 ! 0), 

we obtain sin > — sin 2 -§- 0, 

whence — sin < sin 2 -J- ; 

but sin-J-0 is less than -j-0, and, consequently, sin 2 -J-0 is less than £0 2 ; 

.-. 0-sin0<^. 
4 

(2)' To show that - sin < ^. 

From Problem 11, Set IV., sin 30 = 3sin0 - 4sin 3 0. 

Eeplace successively by 

0_. 

3' 3 2 ' '"' 3 n ' 



TRIGONOMETRIC TABLES. 95 





we obtain 3 sin - — sin = 4 sin 



3sin|-sinH = 4sin 3 |, 



3 sin- sin- — - = 4sm 3 — 

3„ 3„-l 3 n' 

multiplying these equations respectively by 1, 3, 3 2 , • ••, 3 n-1 , and 
adding the results, the following equation is obtained, 

3*sin-^-sin0 = 4/'sin 3 ^ + 3sin 3 ^+ .» +3 w - 1 sin 3 -^Y; 

• 
sin— 

or sin0 = 4 fsin 3 f -f 3 sin 3 4 -I h^sin 8 - 



© 



V 3 3 2 3' 



If the integer n be indefinitely increased, the arc — will tend 
towards 0, and the ratio 

. 
sin — 



3 n 

towards unity ; the first member of the preceding equality has then 
for its limit the difference 6 — sin0, and consequently the second 
member tends also towards this limit. 

But, as the sine is less than its arc, the limit of the second 
member is less than that towards which the geometrical progression 

4 ^ + i 3 + - + ••• + — ^ converges. 
^3 3 3 5 3 rT 3 2n+ V 5 



The limit of this progression = — , consequently 

6 



0-sin0<£ 
6 



96 CONSTRUCTION OF TABLES. 

133. The preceding theorems furnish two limits, that is and 

A3 

, between which sin is comprised. We may easily obtain 

6 

two limits between which cos 6 is comprised. 
We have cos 6 = 1 - 2 sin 2 j 6, 

and, as sin0 is comprised between -J-0 and \Q -, we have at once 

48 

COS0>1-^ 

since cos.<l-2(|-|J,o r <l-| + |_ 2 gj ) 

and, for a stronger reason, 

C os0<l-^ + — . 
2^24 

Therefore cos is comprised between 

1 and I 

2 2 24 



134. The construction of a table of sines and cosines. 

(1) Designate by the length of an arc of 10 seconds. 

Since 7T = 3. 1415926535897932 • • • 

may be taken as the circumference of a circle whose radius equals 
unity, by dividing 10 -k by the number of seconds in a semicircum- 
ference, we obtain 

= ™* = 0.000048481368110 .... 

By Art. 125, sin 6 < 6 and sin 6 > - - ; 

6 

also ^ < 0.000000000000021 ; 

6 

whence sin 10" < 0.000048481368110, 
sin 10" > 0.000048481368089. 



TRIGONOMETRIC TABLES. 97 

These two limits of sin 10" have the first twelve decimal figures 
common, and since they differ by less than half a unit in the 
thirteenth place, we may write 

sin 10"= 0.0000484813681. 

(2) To compute cos 10", we have, Art. 133, 

cos0>l-- and cos0<l-- + -^; 

Jd 2 24: 

and since is < 0.00005 or < 



2.10 4 



4 1 1 

we have — < — — — < 



24 384.10 16 3.10 18 ' 



from which it follows that 1 is a value of cos that differs 

from 1 1 — - by a quantity less than half a unit of the eigh- 

2 24 

teenth decimal order. 

Performing the operation for the first thirteen decimal places, 
we find 

cos 10"= 0.9999999988248. 



135. Sines and cosines of arcs for every 10", up to 45°. 

If, in the formulae 

sin (0 + 0') + sin (0 - 0') = 2 cos 0' sin 0, 
cos (0 -{- 0') + cos(0 - 0') = 2 cos 0' cos 0, 
we put 6 = (m — 1)0', the result will be 



m{ 



sinm0'= 2 cos<9'sin(m - 1)0'- sin(m - 2)0', 
cos m0'= 2 cos 0' cos (m - 1)0'- cos(m - 2)0'. 



If 0' be made equal to 10" and m to 2, then these formulas will 
give the. values of sin 20" and cos 20". Generally, if the sine and 
the cosine of two consecutive multiples of the arc 0'= 10" be known, 
formulas (1) will make known the sine and the cosine of the fol- 
lowing multiple arc. That is to say, Ave may suppose 0' to be 



98 CONSTRUCTION OF TABLES. 

constantly equal to 10", and m to be successively equal to 2, 3, 4, etc. 
We shall then obtain for the sines 

sin 20"= 2cosl0"sinl0"- sin 0"= .0000969627361 

sin 30" =2 cos 10" sin 20" -sin 10" = 

sin40"=2cosl0"sin30"-sin20"= 

Etc., 
and for the cosines 

cos 20"= 2cos 10" cos 10"- cos 0"= 
cos30"=2cosl0"cos20"-cosl0"= 
cos 40"= 2 cos 10" cos 30"- cos 20" = 
Etc. 

The computation may be abridged in the following manner. 
Since the constant multiplier 2 cos 10" differs but little from 2, 

then by placing 2 cos 10"= 2 — k, 

we obtain k = 0.0000000023504. 

The sines may then be written 

sin 20"= 2 sin 10"- sin 0"- fcsin 10", 
sin 30"= 2 sin 20"- sin 10"- k sin 20", 
sin40"=2sin30"-sin20"-ftsin30", etc., 
and the cosines 

cos20"=2cosl0"-cos 0"- k cos 10", 
cos 30"= 2 cos 20" -cos 10" -k cos 20", 
cos 40" = 2 cos 30" - cos 20" - k cos 30", etc. 

This method of computing the sines and cosines of all arcs from 
0" to 30° inclusive, at intervals of 10", is a very simple one. 

136. The sines and cosines of angles or arcs above 30° up to 
and including 45°, are readily obtained by subtraction. 
Remembering that sin 30°= %, we have 

sin(30 o +(9)+sin(30°- 0) = cos0, 
cos(30°- 0) -cos(30°+ 6) = sin 0, 

whence sin (30°+ 0) =cos - sin (50°- 0), 

cos (30° + 6) = cos (30° - 0) - sin 0. 



LOGARITHMS OF TRIGONOMETRIC FUNCTIONS. 99 

If be made equal to 10", 20", 30", etc., successively, then 

sin(30°-f 10") = cos 10" - sin(30°- 10"), 
sin(30° + 20") = cos 20" - sin(30°- 20"), 
sin(30°+30") = cos30" - sin(30°- 30"), etc., 

and cos (30°+ 10") = cos (30°- 10") - sin 10", 

cos(30°+ 20") = cos(30°- 20") - sin 20", 
cos(30°+30") = cos(30°-30")-sin30", etc. 

This process may be continued up to and including 45°. 
The tables for tangents, cotangents, secants, and cosecants can 
be constructed by means of the formulae 

, n sin# A 

tan# = ? sec# = 



cos 6 cos 

, n cos 6 n 1 

cot0 = - — , cosec0 = 



sin sin 

137. The sines and cosines of angles or arcs above 45° will be 
found by taking the cosines and sines of their complements below 
45°, according to the formulae of Art. 17. 

138. The foregoing method is simple in principle, but laborious. 
A much more rapid and simple method is by infinite series. 

TABLES OF LOGARITHMS OF TRIGONOMETRIC FUNCTIONS. 

*139. In numerical applications, computations are very often 
abbreviated by the use of logarithmic tables ; hence there is much 
more need of knowing the logarithms of the sines, cosines, etc., 
of angles or arcs, than the natural functions whose development 
is shown by the preceding Articles. 

If we take the logarithms of the natural sines and cosines of 
all the angles or arcs from 0" to 90° inclusive, another table, called 
the table of logarithmic sines and cosines, will be formed. This 
table once constructed, we may easily form a table of logarithms 
of tangents and cotangents by means of the formulae 

log tan 6 = log sin — log cos 0, 
log cot 6 — log cos 6 — log sin 0. 



100 CONSTRUCTION OF TABLES. 

When desirable the logarithms of secants and cosecants are 
found by means of the formulae 

log cosec 6 = 1 — log cos 0, 
log cosec = 1 — log sin 6. 

140. The logarithms of sines and cosines are never positive 
quantities ; the logarithms of tangents of angles less than 45°, and 
of cotangents of angles greater than 45°, are negative ; therefore, 
to avoid negative quantities in the tables, 10 is added to the 
logarithm of every trigonometric function, thus forming the log- 
arithms of the tables. This increase of the real logarithms by 10 
must always be taken into consideration in logarithmic computations. 

141. Definitions. 

(1) Natural Numbers are arithmetical numbers. 

(2) Tables of Natural Functions are the natural numbers repre- 
senting the values of sines, cosines, etc., when radius is taken equal 
to unity. 

(3) The Logarithm of a natural number is the exponent of the 
power to which another number must be raised to produce the first 
number. 

(4) The Base is the number whose power is to be obtained. 

(5) The base of the Napierian system of logarithms is repre- 
sented by the letter e, whose value = 2.71828 •••. 

(6) The base of the common system is 10. 

(7) A logarithm consists of two parts — characteristic and man- 
tissa. The characteristic is integral, and the mantissa decimal. 

142. Properties of Logarithms. 

(1) The logarithm of a product equals the sum of the logarithms 
of its factors. 

(2) The logarithm of a quotient equals the difference between the 
logarithm of the dividend and that of the divisor. 

(3) The logarithm of any power of a number equals the logarithm 
of the number multiplied by the exponent of the poiver. 

(4) The logarithm of any root of a number equals the logarithm 
of the number divided by the index of the root. 



LOGARITHMS OF TRIGONOMETRIC FUNCTIONS. 101 

143. The Arithmetical Complement of a logarithm is the re- 
mainder after subtracting the logarithm from zero. 

The arithmetical complement, or co-log, as it is frequently called, 
is used when addition is substituted for subtraction, on the prin- 
ciple that adding the co-logarithm of a number is the same precisely 
as subtracting the logarithm. 

144. The logarithmic and trigonometric tables that may be 
consulted present some variety in their mode of arrangement, and 
are usually accompanied with full explanation of their peculiarities 
and the methods of using the tables. It is not necessary to enter 
into any minute account of the way in which tables may be used 
with the greatest advantage. The student is referred to the explana- 
tions accompanying the tables to be used. 



APPENDIX. 



The formulae of the preceding pages are of great importance. 
Collected and numbered the same as in the text, they will be found 
convenient for reference and use. 



Page 6. 



PLANE TRIGONOMETRY. 

Trigonometric ratios. 



(1) 


sin^ = l% 
c 


(2) 


cos-4 = -, 
c 


(3) 


6 


W 


cot A = -, 
a 


(5) 


sec-4 = - 5 


(6) 


cosec^d = -, 



(7) vers J. = 1 — cos A, 

(8) covers ^4 = 1 — sin ^L 



[i] 



(1) 


tan^l x cotJ. = l, 


(2) 


sin^l x cosec.4 = 1, 


(3) 


cos.4x seo^ = l, 


(4) 


taiul= sin ^. 
cos A 


(5) 


otA = cosA - 



sin A 



[2] 



103 



104 



APPENDIX. 



Page 7. 



Page 8. 



(1) 
(2) 
(3) 

(5) 
(6) 

(1) siivU + cos 2 ^. = 1, 



a a 

smi=- = 

c 


cos 5, 


a h 
cos A = - = 

c 


sin 5, 


tan A = - = 
b 


cot 5, 


a 


tan_B, 


sec J. = - = 
6 


cosec_B, 


cosecJ. =-■= 
a 


sec B. 



(2) 
(3) 
(4) 

(5) 

(6) 

(7) 
(8) 



1 + ts.tr A = secl4, 
1 + cot 2 ^4 = cosecM, 



COS .4 = 

tan^l = 


VI - sin 2 ^, 
sin A 


Vi 


5 

— sin 2 A 


cot A = 

sec A = 
cosec A = 


vr 


— sin 2 ^4 


sin A 
1 


Vi 
1 


— sinlA 



sm -4 



[3] 



M 



Functions of the sum and the difference of two angles or arcs. 

Page 24. 

sin (0 + 0') = sin cos 0' + cos sin 0'. 

cos (0 + 0') = cos cos 0'— sin sin 0'. 
sin (0 - 0') = sin cos 0'- cos sin 0'. 
cos (0 - 0') = cos cos 0' + sin sin 0'. 

tan(0 + 0')= J^' + tany 

v y l-tan0tan0' 



[5] 
[6] 

[8] 



APPENDIX. 



105 



Page 25. 



cot (0 + 0') = 
tan(0-0') = 
cot (0 - 0') = 



cot cot 



cot 0' + cot 


tan0-tan0' 


l-f-tan0tan0' 


l + cot0cot0' 



cot $' -cote 



[10] 

[11] 
[12] 



Functions of the sum of three angles or arcs. 
sin (0 + 0' + 0") = sin0 cos 0' cos 0"-f- sin0'cos0 cos0" 



-f sin0"cos0cos0' — sin (9 sin0'sin0". 

cos (0 4- 0'+ 0") = cos cos 0' cos 0"- cos sin 0' sin 0" ] 

( 
J 



^ [13] 



Page 26. 



sin(0 + 0') + sin(0 - 0') = 2 sin0 cos 0', 
sin(0 + 0') - sin(0 - f ) = 2 cos sin 0'. 

cos (0 + 0') + cos (0 - 0') = 2 cos cos 0', 
cos(0 + 0') - cos(0 - 0') = - 2sin0 sin0'. 



[14] 



y [is] 



Functions of the sum and the difference of the sines and cosines of 

two angles. 



(1) sin<£ +sin<£ r =2sini(<£ + <£')cos!(</> -<£'), 

(2) sin 4> - sin <j>'= 2sini(cf> - <f>')cos±(cf> + <£'), 

(3) cos<£ + cos<£'=2cosi(c£4-<£')cos£(<£ — <£'), 

(4) cos<£ — cos<£'=2sin-J(<£ + (£')sin-|-(</> — <£'). 



cos^ + sin^=2sin^7r-^^V os ri 7J "-^^\ 



cos</> — sin<fr f =2sin( ^tt — ^ - 



)-(*-***) 



[16] 



[17] 



106 



APPENDIX. 



Page 27. 
^v sin <fr + sin 4> 



sin <f> — sin <£ 
/0 \ sin <£ + sin <f> 



cos <£ + cos <£ 
,o\ sin j» + sin 



Combinations formed from [16]. 

. sin£(<fr+<fr f )cos£(<fr— <ftO = taii£(<ft+<ft') 
"sin£(«£-<£')cosi(<£+0') tan£(<£-<£') 

.sinj-Q+cft') 



cos </> — COS <£ 
,.< sin eft — sin <fr 



cos<£ + cos</> 
.k\ sin<fr f — sin <fr 



cos c£ — cos <£ 

,n^ COS<£ -h COS <£ 
COS<£' — COS<£ 



cosi(<H-<£') 

= cosj-(>-^) 
sin | (<£-<£') 

cos|(</>-^> r ) 
= cosK<fr+(fr') 

= cos £(<£ + <£') CQS^(^)-^ f ) 
sinj(^+^')sinj(^-^) 

=coti(++^')ooti(*-^)- 



Functions of double angles. 
sin 2 = 2 sin cos 0. 
cos 26 — cos 2 — sin 2 0. 
2tan0 



= tan £(<£ + <£'), 

= cbti(*-*')> 

= tan ■§-(<£— <£'), 

= cotj-(>-}-<£') ? 



[18] 



tan2(9 = 
cot 2(9 = 



l-tan 2 

cot 2 0-l 
2cot0 



[19] 
[20] 

[21] 
[22] 



Page 28. 



Functions of half angles. 



sin£0 = ± i( VI + sin0 qp Vl-sin0). 



cos|0 = ± i( Vl + sin0 ± VI- sin 0). 



cos 



sin£0 = ±^i 



— COS0 



[23] 
[24] 

[25] 

[26] 



APPENDIX. 










107 


tan i0_ sin# _1- C os0 










[27] 


1 + costf sing 


sin 1 — cos 


cos 








[28] 


sin20 = ± 2sin(9Vi - sin 2 <9 = ± 2 


0V1- 


-cos 1 


e. 


[29] 


cos20 = 1 - 2sin 2 = 2cos 2 # - 1. 










[30] 



Page 31. 



Page 32. 



Laiv of sines. 



smA sin B sin G 

A+B+C= 180°, 

a b 

sin ^4 sin I?' 

a c 



sin ^1 sin G 



[311 



[32] 



Xcwo of cosines. 

c 2 = a 2 + b 2 -2abcosC, 
a 2 =6 2 + c 2 - 2bc cos A 



79 

o- = a- 



2 ac cos B. 



[33] 



££c?e o/ a triangle in terms of the cosines of the adjacent angles emd? 
the other two sides. 
Page 33. 

(1) a = b cos C + c cos B, T 

(2) b = c cos A + a cos (7, > [34] 

(3) c = acosB + bcosA. . 



Page 34. 



Zcat; of tangents. 

a + 6 == tanj-(^+^) 
a— 6 tan-J-(^l— J3) 



[35] 



108 



APPENDIX. 



Functions of half angles in terms of the sides of a triangle. 
Page 35. 



Page 36. 



Page 37. 



suUi = 



* ac 

\ ah 



sin 



sin J 



eos i5 = 



-V 



S (6' - b) 



cos 



J 8( ,_ c) > 
2 \ ab 



2 \ s (s - a 



) ' 



tan I- B 



l (s-a)(s-c) 
A s (s - b) ' 



* S (.9 — 



-g)( s-&) 

<0 



Area of triangles. 
K=\bc$mA. 

K = Vs (s - a) (s -&)(«- c) . 



[36] 



[37] 



[38] 



[39] 
[40] 



(1) sin -J- J. sin J- B sin -J- C = 



(s — a) (g — b) (s — c) 
abc 



K 2 

sabc 



/o\ t a ^ -n \ n s Vs(S — a)(S — b)(s — c) As i FAil 

(2) coslAcoslBcos^C = * ^ ^ ^ = — , ? L 41 J 

a 6c aoc 



(3) tan±^4taniJ3tan|C ■ 



APPENDIX. 



109 



Radii of circumscribed and inscribed circles. 



Page 38. 



R = 



abc 



4 Vs(s — a) (s — b)(s — c) 
r = J p-«)p-fr)Q-c) 



[42] 



[43] 



Page 39. 



Page 40. 



Radii of escribed circles. 

4 



s — a 






K 

s-b 



!s(s 


-b)(s- 


-c) 


s — a 


s(s- 


-a)(s- 


-b) 


s — c 


s(s 


-a){s- 


-o) 



r' =stan-J-^l, 
r" = stan 1(7, 
r'"=s tan \ B. 



s-b 
1 



l = l . : 



.+*+* i 



ir=V 



?•?•'■>•"?• 



45*r»+-,-!'+r'"-f. 



SPHEETCAL TRIGONOMETRY. 



Fundamental formulae, of spherical triangles. 



Page 51. 



cos a = cos 6 cos c -f- sin & sin c cos ^4, 
cos 6 = cos a cos c + sin a sin c cos B, 
cos c = cos a cos 6 -j- sin a sin 6 cos C. 



[44] 



[45] 



[46] 



[47] 



110 



APPENDIX. 



Page 52. 



cos A = — cos B cos C 4- sin B sin C cos a, 
cos B = — cos ul cos (7 + sin A sin C cos 6, 
cos C= — cos ^L cos 2> + sin A sin _B cos c. 



[48] 



Page 53. 



Laiv of sines, 
sin A smB sin<7 



sm a 



sin 6 



sine 



[49] 



Three sides and two angles of a spherical triangle. 

Page 54. 

cos a sin b — sin a cos b cos (7 = sin c cos ^1, 
'COS-6 sin a — sin 6 cos a cos = sin c cos 5, 
•cos-6 sin e — sin & cos € cos ^4 = sin a cos jB, 
cose sin b — sine cos 6 cos A = sin a cos C, 
cos e sin a — sin c cos a cos B = sin 6 cos G, 
cos a sin c — sin a cos c cos J5 = sin b cos A 



[50] 



Titfo sides and three angles of a spherical triangle. 

cos a sin B — cos 6 cos C sin ^4 = cos A sin (7, 
■cos b sin ^L — cos a cos (7 sin jB = cos B sin (7, 
cos b sin (7 — cos c cos ^4 sin B = cos jB sin A, 
cos c sin 5 — cos b cos ^L sin O = cos (7 sin A, 
cos c sin A — cos a cos B sin (7 = cos C sin 5, 
cos a sin C — cos c cos B sin ^4 = cos A sin 2?. 



[51] 



Tivo sides and two angles of a spherical triangle. 

cot a sin b — cot A sin (7 = cos b cos (7, 
cot & sin a — cot 5 sin C = cos a cos C, 
cot & sine — cot B sin A = cose cos ;4, 
cote sin 5 —cot (7 sin J. = cos 6 cos A, 
cot c sin a — cot (7 sin B = cos a cos B, 
cot a sine — cot A sin B = cos c cos B. 



[52] 



APPENDIX. 



Ill 



Formulce for 
Page do. ,., 

(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(3) 
(9) 
(10) 



solving spherical right triangles. 

cos a = cos b cose, 
sin b = sin a sin.23, 
sin c = sin a sin C, 
tan b = tan a cos (7, 
tan & = sin c tan 5, 
tan c = -tan a cos B, 
tan c = sin b tan (7, 
cos a = cot-BcotO, 
cos B= cosh sin C, 
cos (7= cose sin B. 



[53] 



Functions of half angles in terms of the sides of a spherical triangle. 

Page 56. 

smJ?A 



/ sin(s — 5)sin(s — c) 



sin b sin e 



sinJr£ 



_ / sin (s — a) sip (s — c) 
\ sin a sin c 



sini(7 = 



/sin (s — a) sin (.s — & ) 
A sin a sin b 



[54] 



cos J^.= 
cos -J- 12 = 
cos hC = 
tan-^ = 
tan ±B = 
tani<7 = 



/sing sin(s — a) 
\ sin b sin c 



'sin s sin(s — 5) 
sin a sine 



,'sins sin(s — c) 
\ sin a sin 6 

/sin (s — 6) sin (s — c) 
v sin s sin (s — a) 

/sin (s — a)sin(s — c) 
V sins sin (s — 5) 

/sin(s — a)sin(s — b) 
v sins sin (s — c) 



[55] 



[56] 



112 



APPENDIX. 



Functions of half sides in terms of the angles of a spherical triangle. 
Page 58. 



sin -k a = 



cosS cos (S —A) 



sinB sin C 



sin -J b 



-v 



cosff cos (& — B) 
sin A sinC 



. , /' cos S cos (S-C) 

\ sin ^4 sin I? 



[57] 



cos 



/ COS(£-B) COS(S-C) 

\ sin 5 sin C 



cos -J- 6 



cos(^--^L) cos(ff-O) 
\ sin A sin (7 



cos 



*c=V 



cos(^-^L) cos (S-B) 
sin A sinB 



[58] 



tan 



tan 






cosS cos (S — A) 
cos (S-B) cos (£-0)' 



P=V~ 



eos<S cos (<S — B) 
cos (.5 — A) cos (S—C) 

cos S cos (S - 6) 
\ cos (S — A) oos (S-B) 



[59] 



Page 60. 



(1) 
(2) 
(3) 



Delambre's Formulae, 



sin j-(yl + 5) _ cosj-(a — b) 



cos^C 


cos^-c 


sin±(A-B) 


_ sin|-(a — b) 


cos^C 


sin J c 


cosi(A-hB) 


_ cos -J- (a + &) 


sin \G 


cos-^-c 


cosi(A-B) 


sin|-(a + b) 



sin A- (7 



sin -k c 



[60] 



APPENDIX. 



113 



Napier's Analogies. 



(1) tan^ + g) = coHC COS ^ a ~^ 



(2) tan-£(^-JB) = cot£<7 



sin ^(a — b) 
sin£(a + &) 



(3) tani(a + 6) =tan|c G0S i( A — ^ 

(4) tan±(a-6) =tan^c sm i(^-^) 
v ; 2V J 2 sini(^. + ^) 



[61] 



Page 76. 



Area of spherical triangles. 



ABG = § X^. 



[62] 



VHuillier's Formula. 



Page 77. 






tan^i? = Vtan-^-s tan^-(s — a) tan^-(s — b) tan|-(s — c). [63] 



u.4rea o/ spherical polygons. 
P= [T-{n-2)7ry. 



[64] 



Angular radii of small circles circumscribed about, inscribed in, and 
escribed upon, the sides of spherical triangles. 



Page 79. 



tanj8 = 



tan-S-c 



sin(0-iJS7) 



[65] 



tan r 



/ sin(s — a) sin(s — b) sin(s — c) 
\ sin s 



[66] 



114 



APPENDIX, 



Page 80. 

tan r' 



V sin(g — b) sin(g — c) _ / sing sin ( g — b) si 
sin s sin (s — a) \ sin (g — a 



sin (g — c) 

) " ! 



tanr 



Vsin(g — a) sin(g — c) /sing sin(g — a) sin(g — c) 
. . . __ -* / : . , 
sin a sin (s — h~\ \ sm(g — 6) 



[67] 



sing sin(g — b) 



tan r 



= sing a/ 



sin(g — a) sin(g — b) 
sins sin (g — c) 



Vsing sin(g — a) sin( 
sin (g — c) 



s-b) 



LIBRARY OF CONGRESS 

1 llllllilllllllllll 



mil lam fiiiii liiii IlIiE ilJII lEIll liiil IIIM 11 

O 003 560 697 A 




mP 



